<s>[INST] Proving $\sum_{k=0}^{m}\binom{n+k}{k}(1-x)^kx^{n+1}=1-\sum_{k=0}^{n}\binom{m+k}{k}x^k(1-x)^{m+1}$ without calculus How would you prove this without calculus?

$$\forall m,n\in\Bbb N,\ \forall x\in\Bbb R,\ \sum_{k=0}^{m}\binom{n+k}{k}(1-x)^kx^{n+1}=1-\sum_{k=0}^{n}\binom{m+k}{k}x^k(1-x)^{m+1}$$

In this post it is proved with calculus, and I was wondering if there was an easier way to prove it. I tried proving it with induction, by holding $n$ constant, however I didn't seem to get anywhere.
 [/INST] Induction works.
Let
$$a(n,m)=x^{n+1}\sum_{k=0}^{m}\binom{n+k}{k}(1-x)^k$$
$$b(n,m)=(1-x)^{m+1}\sum_{k=0}^{n}\binom{m+k}{k}x^k$$
$$f(n,m)=a(n,m)+b(n,m)$$
So we have to prove $f(n,m)=1$ for all $n,m\ge0$.
Let's prove this by induction on $m$.
First, $f(n,0)=1$ for all $n\ge0$, as $a(n,0)=x^{n+1}$ and $b(n,0)=(1-x)\sum_{k=0}^nx^k=1-x^{n+1}$.
Below, $n\ge0$ is fixed.
Let's assume $f(n,m)=1$ for some $m\ge0$. Then:
$$a(n,m+1)=a(n,m)+x^{n+1}\binom{n+m+1}{m+1}(1-x)^{m+1}$$
$$b(n,m+1)=(1-x)^{m+2}\sum_{k=0}^n\binom{m+k+1}{k}x^k=(1-x)^{m+2}\left\{\sum_{k=0}^n\binom{m+k}{k}x^k+\sum_{k=1}^n\binom{m+k}{k-1}x^k\right\}\\=(1-x)(1-x)^{m+1}\sum_{k=0}^n\binom{m+k}{k}x^k+(1-x)^{m+2}\sum_{k=1}^n\binom{m+k}{k-1}x^k\\=(1-x)b(n,m)+(1-x)^{m+2}\sum_{k=0}^{n-1}\binom{m+k+1}{k}x^{k+1}$$
Therefore
$$f(n,m+1)=f(n,m)+(1-x)^{m+1}\left\{\binom{n+m+1}{m+1}x^{n+1}-x\sum_{k=0}^n\binom{m+k}{k}x^k+(1-x)\sum_{k=0}^{n-1}\binom{m+k+1}{k}x^{k+1}\right\}$$
We want to prove that the expression inside the curly brackets is zero:
$$\binom{n+m+1}{m+1}x^{n+1}-x\sum_{k=0}^n\binom{m+k}{k}x^k+(1-x)\sum_{k=0}^{n-1}\binom{m+k+1}{k}x^{k+1}\\=\sum_{k=0}^n\binom{m+k+1}{k}x^{k+1}-x\sum_{k=0}^n\binom{m+k}{k}x^k-x\sum_{k=0}^{n-1}\binom{m+k+1}{k}x^{k+1}\\
=\sum_{k=0}^n\binom{m+k+1}{k}x^{k+1}-x\sum_{k=0}^n\binom{m+k}{k}x^k-x\sum_{k=1}^{n}\binom{m+k}{k-1}x^{k}\\
=\sum_{k=0}^n\binom{m+k+1}{k}x^{k+1}-\sum_{k=0}^n\binom{m+k}{k}x^{k+1}-\sum_{k=1}^{n}\binom{m+k}{k-1}x^{k+1}\\
=x-x+\sum_{k=1}^n\left\{\binom{m+k+1}{k}-\binom{m+k}{k}-\binom{m+k}{k-1}\right\}x^{k+1}\\=0$$
Where the $x-x$ in the last line comes from the index $k=0$ in the first two sums of the preceding line.
Therefore the induction step is proved: if $f(n,m)=1$, then $f(n,m+1)=1$. And by induction, the equality is true for all $m\ge0$. Since this is proved for any $n\ge0$, the equality $f(n,m)=1$ is true for all $n,m\ge0$.
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