<s>[INST] Exotic proofs of $\sum_{j=0}^{n-1}\binom{p+j}{p}=\binom{p+n}{p+1}$ Let $p,n$ be positive integers.
The following identity $\displaystyle \sum_{j=0}^{n-1}\binom{p+j}{p}=\binom{p+n}{p+1}$ may be proved by induction or by successive uses of Pascal's rule (both techniques are demonstrated here).
Question
Is there a combinatorial proof for this identity ? Do you know other proofs ?
 [/INST] First Proof:
\begin{align}
\sum^{n-1}_{j=0}\binom{p+j}{p}
&=\frac{1}{2\pi i}\oint_{|z|=1}\frac{(1+z)^p}{z^{p+1}}\sum^{n-1}_{j=0}(1+z)^j{\rm d}z\\
&=\frac{1}{2\pi i}\oint_{|z|=1}\frac{(1+z)^{n+p}-(1+z)^p}{z^{p+2}}{\rm d}z\\
&=\frac{1}{(p+1)!}\lim_{z \to 0}\frac{{\rm d}^{p+1}}{{\rm d}z^{p+1}}\left[(1+z)^{n+p}-(1+z)^p\right]\\
&=\frac{(n+p)(n+p-1)\cdots(n+1)(n)}{(p+1)!}\\
&=\binom{n+p}{p+1}
\end{align}
Second Proof:
\begin{align}
\sum^{n-1}_{j=0}\binom{p+j}{p}
&=[x^N]\sum^\infty_{N=0}\sum^{N=n-1}_{j=0}\binom{p+j}{j}x^N\\
&=[x^N]\sum^\infty_{j=0}\binom{p+j}{j}\sum^\infty_{N=j}x^N\\
&=[x^N]\sum^\infty_{j=0}\binom{p+j}{j}\frac{x^j}{1-x}\\
&=[x^{n-1}]\frac{1}{(1-x)^{p+2}}\\
&=\binom{p+2+n-1-1}{n-1}\\
&=\binom{p+n}{p+1}
\end{align}
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