Let $n\geq 0$,and suppose that $x_i$,$i=0,\cdots,n$ are distinct real numbers.Then,given two sets of real numbers $y_i,i=0,\cdots,n$,and $z_i,i=0,\cdots,n$,there is a unique polynomial $p_{2n+1}$ in $\mathcal{P}_{2n+1}$such that
\begin{equation}
p_{2n+1}(x_i)=y_i,p'_{2n+1}(x_i)=z_i,i=0,\cdots,n
\end{equation}
Proof:Let the polynomial be in the form of
\begin{equation}
a_{2n+1}x^{2n+1}+a_{2n}x^{2n}+\cdots+a_1x+a_0
\end{equation}
Then
\begin{align*}
\begin{cases}
a_{2n+1}x_0^{2n+1}+a_{2n}x_0^{2n}+\cdots+a_1x_0+a_0=y_0\\
\vdots\\
a_{2n+1}x_n^{2n+1}+a_{2n}x_n^{2n}+\cdots+a_1x_n+a_0=y_0\\
\end{cases}
\end{align*}
And
\begin{align*}
\begin{cases}
(2n+1)a_{2n+1}x_0^{2n}+2na_{2n}x_0^{2n-1}+\cdots+a_1+0\cdot a_0=z_0\\
\vdots\\
(2n+1)a_{2n+1}x_n^{2n}+2na_{2n}x_n^{2n-1}+\cdots+a_1+0\cdot a_0=z_n\\
\end{cases}
\end{align*}
Now we see the determinant
\begin{equation}
\begin{vmatrix}
x_0^{2n+1}&x_0^{2n}&\cdots&x_0&1\\
x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\
\vdots&\vdots&\cdots&\vdots\\
x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\
(2n+1)x_0^{2n}&2nx_0^{2n-1}&\cdots&1&0\\
\vdots&\vdots&\cdots&\vdots&\cdots\\
(2n+1)x_n^{2n}&2nx_n^{2n-1}&\cdots&1&0\\
\end{vmatrix}
\end{equation}
Now we prove that this determinant is nonzero.We first study some concret examples.
When $n=0$,the determinant is
\begin{equation}
\det \begin{pmatrix}
x_0^1&1\\
1&0\\
\end{pmatrix}
\end{equation}
Which is equal to $-1$.When $n=1$,the determinant is
\begin{equation}
\det \begin{pmatrix}
x_0^3&x_0^2&x_0&1\\
x_1^3&x_1^2&x_1&1\\
3x_0^2&2x_0&1&0\\
3x_1^2&2x_1&1&0\\
\end{pmatrix}
\end{equation}
This determinant is equal to
\begin{equation}
\det \begin{pmatrix}
x_0&1&x_0^3&x_0^2\\
x_1&1&x_1^3&x_1^2\\
1&0&3x_0^2&2x_0\\
1&0&3x_1^2&2x_1\\
\end{pmatrix}
\end{equation}
We know that
\begin{equation}
\begin{pmatrix}
x_0&1\\
x_1&1\\
\end{pmatrix}
\end{equation}
is invertible,so
\begin{align*}
\det \begin{pmatrix}
x_0&1&x_0^3&x_0^2\\
x_1&1&x_1^3&x_1^2\\
1&0&3x_0^2&2x_0\\
1&0&3x_1^2&2x_1\\
\end{pmatrix}=\det \begin{pmatrix}
x_0&1\\
x_1&1\\
\end{pmatrix}\det \left[ \begin{pmatrix}
3x_0^2&2x_0\\
3x_1^2&2x_1\\
\end{pmatrix}-\begin{pmatrix}
1&0\\
1&0\\
\end{pmatrix}\begin{pmatrix}
x_0&1\\
x_1&1\\
\end{pmatrix}^{-1}\begin{pmatrix}
x_0^3&x_0^2\\
x_1^3&x_1^2\\
\end{pmatrix}\right]
\end{align*}
\begin{equation}
\begin{pmatrix}
x_0&1\\
x_1&1\\
\end{pmatrix}^{-1}=\frac{1}{x_0-x_1}\begin{pmatrix}
1&-1\\
-x_1&x_0\\
\end{pmatrix}
\end{equation}
\begin{equation}
\begin{pmatrix}
1&0\\
1&0\\
\end{pmatrix}\begin{pmatrix}
1&-1\\
-x_1&x_0\\
\end{pmatrix}=\begin{pmatrix}
1&-1\\
1&-1\\
\end{pmatrix}
\end{equation}
\begin{equation}
\begin{pmatrix}
1&-1\\
1&-1\\
\end{pmatrix} \begin{pmatrix}
x_0^3&x_0^2\\
x_1^3&x_1^2\\
\end{pmatrix}=\begin{pmatrix}
x_0^3-x_1^3&x_0^2-x_1^2\\
x_0^3-x_1^3&x_0^2-x_1^2\\
\end{pmatrix}
\end{equation}
so we just need to consider
\begin{equation}
\begin{pmatrix}
3x_0^2&2x_0\\
3x_1^2&2x_1\\
\end{pmatrix}-\begin{pmatrix}
x_0^2+x_1^2+x_0x_1&x_0+x_1\\
x_0^2+x_1^2+x_0x_1&x_0+x_1\\
\end{pmatrix}=\begin{pmatrix}
2x_0^2-x_1^2-x_0x_1&x_0-x_1\\
2x_1^2-x_0^2-x_0x_1&x_1-x_0\\
\end{pmatrix}
\end{equation}
so we just need to consider
\begin{equation}
\det \begin{pmatrix}
2x_0^2-x_1^2-x_0x_1&x_0-x_1\\
2x_1^2-x_0^2-x_0x_1&x_1-x_0\\
\end{pmatrix}=(x_0-x_1)^2(x_1-x_0)
\end{equation}
So
\begin{equation}
\det \begin{pmatrix}
x_0^3&x_0^2&x_0&1\\
x_1^3&x_1^2&x_1&1\\
3x_0^2&2x_0&1&0\\
3x_1^2&2x_1&1&0\\
\end{pmatrix}=-(x_0-x_1)^4
\end{equation}
When $n=2$,the determinant is
\begin{equation}
\det\begin{pmatrix}
x_0^5&x_0^4&x_0^3&x_0^2&x_0&1\\
x_1^5&x_1^4&x_1^3&x_1^2&x_1&1\\
x_2^5&x_2^4&x_2^3&x_2^2&x_2&1\\
5x_0^4&4x_0^3&3x_0^2&2x_0&1&0\\
5x_1^4&4x_1^3&3x_1^2&2x_1&1&0\\
5x_2^4&4x_2^3&3x_2^2&2x_2&1&0\\
\end{pmatrix}
\end{equation}
By using mathematica,it is easy to see that
\begin{equation}
\det\begin{pmatrix}
x_0^5&x_0^4&x_0^3&x_0^2&x_0&1\\
x_1^5&x_1^4&x_1^3&x_1^2&x_1&1\\
x_2^5&x_2^4&x_2^3&x_2^2&x_2&1\\
5x_0^4&4x_0^3&3x_0^2&2x_0&1&0\\
5x_1^4&4x_1^3&3x_1^2&2x_1&1&0\\
5x_2^4&4x_2^3&3x_2^2&2x_2&1&0\\
\end{pmatrix}=(x_0-x_1)^4(x_0-x_2)^4(x_1-x_2)^4
\end{equation}
Now we prove that
\begin{equation}
\begin{vmatrix}
x_0^{2n+1}&x_0^{2n}&\cdots&x_0&1\\
x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\
\vdots&\vdots&\cdots&\vdots\\
x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\
(2n+1)x_0^{2n}&2nx_0^{2n-1}&\cdots&1&0\\
\vdots&\vdots&\cdots&\vdots&\cdots\\
(2n+1)x_n^{2n}&2nx_n^{2n-1}&\cdots&1&0\\
\end{vmatrix}=(-1)^{n}\prod_{n\geq i>j\geq 0}(x_i-x_j)^4
\end{equation}
Let
\begin{equation}
f(x)=
\begin{vmatrix}
x^{2n+1}&x^{2n}&\cdots&x&1\\
x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\
\vdots&\vdots&\cdots&\vdots\\
x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\
(2n+1)x^{2n}&2nx^{2n-1}&\cdots&1&0\\
\vdots&\vdots&\cdots&\vdots&\cdots\\
(2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\
\end{vmatrix}
\end{equation}
It is easy to verify that $f(x)$ is a polynomial of degree $4n$.And
\begin{equation}
f(x_1)=\cdots =f(x_n)=0
\end{equation}
So $(x-x_1)(x-x_2)\cdots (x-x_n)|f(x)$.And
\begin{align*}
f'(x)= \begin{vmatrix}
x^{2n+1}&x^{2n}&\cdots&x&1\\
x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\
\vdots&\vdots&\cdots&\vdots\\
x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\
2n(2n+1)x^{2n-1}&(2n-1)2nx^{2n-2}&\cdots&0&0\\
\vdots&\vdots&\cdots&\vdots&\cdots\\
(2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\
\end{vmatrix}
\end{align*}
So $f'(x_1)=\cdots f'(x_n)=0$.So $(x-x_1)^2(x-x_2)^2\cdots
(x-x_n)^2|f(x)$.And
\begin{align*}
f''(x)= \begin{vmatrix}
(2n+1)x^{2n}&2nx^{2n-1}&\cdots&1&0\\
x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\
\vdots&\vdots&\cdots&\vdots\\
x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\
2n(2n+1)x^{2n-1}&(2n-1)2nx^{2n-2}&\cdots&0&0\\
\vdots&\vdots&\cdots&\vdots&\cdots\\
(2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\
\end{vmatrix}+
\begin{vmatrix}
x^{2n+1}&x^{2n}&\cdots&x&1\\
x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\
\vdots&\vdots&\cdots&\vdots\\
x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\
(2n-1)2n(2n+1)x^{2n-2}&(2n-2)(2n-1)2nx^{2n-3}&\cdots&0&0\\
\vdots&\vdots&\cdots&\vdots&\cdots\\
(2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\
\end{vmatrix}
\end{align*}
It is easy to verify that $f''(x_1)=\cdots f''(x_n)=0$.So
\begin{equation}
(x-x_1)^3(x-x_2)^3\cdots (x-x_n)^3|f(x)
\end{equation}
And it is also easy to figure out $f'''(x)$,so it is easy to verify that
\begin{equation}
f'''(x_1)=\cdots =f'''(x_n)=0
\end{equation}
So
\begin{equation}
(x-x_1)^4(x-x_2)^4\cdots (x-x_n)^4|f(x)
\end{equation}
Because $f(x)$ is a polynomial of degree $4n$,so $f(x)=a(x-x_1)^4(x-x_2)^4\cdots (x-x_n)^4$.According to symmetry ammong $x_0,x_1,\cdots,x_n$ in this determinant,it is easy to verify that
\begin{equation}
f(x_0)=c\prod_{n\geq i>j\geq 0}(x_i-x_j)^4
\end{equation}
Then we prove that $c=(-1)^n$.We do it by induction.It is easy to verify that When $n=1$,\begin{equation} \det \begin{pmatrix} x_0^3&x_0^2&x_0&1\\ x_1^3&x_1^2&x_1&1\\ 3x_0^2&2x_0&1&0\\ 3x_1^2&2x_1&1&0\\ \end{pmatrix}=-(x_0-x_1)^4 \end{equation}Then when n=2,Let's see the determinant \begin{equation} \det\begin{pmatrix} x_0^5&x_0^4&x_0^3&x_0^2&x_0&1\\ x_1^5&x_1^4&*x_1^3&*x_1^2&*x_1&*1\\ x_2^5&x_2^4&*x_2^3&*x_2^2&*x_2&*1\\ 5x_0^4&4x_0^3&3x_0^2&2x_0&1&0\\ 5x_1^4&4x_1^3&*3x_1^2&*2x_1&*1&*0\\ 5x_2^4&4x_2^3&*3x_2^2&*2x_2&*1&*0\\ \end{pmatrix} \end{equation} The element marked * also form a determinant,we know that the constant of this determinant is -1,so the constant term of the determinant is $-1\times (4-5)=1$.……So by induction we know that $c=(-1)^n$.
So the determinant
\begin{equation}
\begin{vmatrix}
x_0^{2n+1}&x_0^{2n}&\cdots&x_0&1\\
x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\
\vdots&\vdots&\cdots&\vdots\\
x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\
(2n+1)x_0^{2n}&2nx_0^{2n-1}&\cdots&1&0\\
\vdots&\vdots&\cdots&\vdots&\cdots\\
(2n+1)x_n^{2n}&2nx_n^{2n-1}&\cdots&1&0\\
\end{vmatrix}
\end{equation}
is nonzero,so $p_{2n+1}$ exists and unique.
Remark :There is some discussion to this problem in Prove an identity including determinant