Combinatorial identity and Fuss-Catalan numbers I would like to show that
$$
\lim_{N\to\infty}\frac{1}{N^{np+1}}\frac1{p!}\sum_{j=0}^{p-1}(-1)^j\binom{p-1}{j}
\left(\frac{\Gamma(N+p-j)}{\Gamma(N-j)}\right)^{n+1}
=\frac1{np+1}\binom{(n+1)p}{p},
$$
for $p,n=1,2,\ldots$.
Background
The reason we expect this equality to hold is the following: The left hand side appear in [arXiv:1307.7560] as the moments of the squared singular values of a product of $n$ Gaussian non-Hermitian $N\times N$ matrices, while the right hand side (Fuss-Catalan numbers) is the asymptotic prediction obtained using techniques from free probability, see e.g. [arXiv:0710.5931]. Thus the results should agree at leading order in $N$.
 A: Note that $$
f(N):=\left(\frac{\Gamma(N+p)}{\Gamma(N)}\right)^{n+1}
$$
is a unitary polynomial in $N$ of degree $pn+p$: $f(N)=N^{pn+p}+\dots$. Its $(p-1)$-st finite difference 
$$
\Delta^{p-1}f(N)=\sum_{j=0}^{p-1} (-1)^j\binom{p-1}{j} f(N-j)
$$
is a polynomial in $N$ of degree $(pn+p)-(p-1)=pn+1$ and leading coefficient $(pn+p)(pn+p-1)\dots(pn+p-(p-2))$. It yields your limit relation.  
A: From taking derivatives of $(x-1)^{p-1}$ and setting $x = 1$, we have
$$\sum_{i=0}^{p-1} j^k (-1)^j \binom{p-1}{j} = 0$$
for $k < p - 1$. The same method gives
$$(-1)^{p-1}(p- 1)! = \sum_{i=0}^{p-1} j^{p-1} (-1)^j \binom{p-1}{j}.$$
Now consider $$\left(\frac{\Gamma(N+ p - j)}{\Gamma(N - j)}\right)^{n+1} = (N + p - j - 1)^{n+1}\cdot \dots \cdot(N - j)^{n+1}.$$ This can be considered a polynomial in two variables: $j$ and $N$. Consider a term $Kj^aN^b$ in the decomposition into monomials. If $a < p - 1$, then the term will cancel via the above identity. If $b < np + 1$, then the term will contribute a negligible amount asymptotically. Then we can focus on $a = p - 1$, $b = np + 1$. But the $j^{p-1}N^{np+1}$ term of this polynomial will equal the $j^{p-1}N^{np+1}$ term of the homogenous $(N - j)^{(n+1)p}$. The coefficient of the term is $(-1)^{p-1}\binom{(n+1)p}{np+1}$. Combining with the above identity, we find that the left hand side is
$$\frac{(p-1)!}{p!} \binom{(n+1)p}{np+1} + O(N^{-1}) = \frac{1}{np+1} \binom{(n+1)p}{p} + O(N^{-1})$$
which gives you what you need.
