Closed form summation of expression with binomial coefficient I am trying to find a closed form expression of the sum below:
$$ \mathbb{E}(S) = \sum_{s=0}^{N-n} s {N-s-1 \choose n-1} $$
I have considered summation by parts
$$ \sum_{k=m}^n f_k \Delta g_k = f_ng_{n+1} - f_mg_m - \sum_{k=m}^{n-1}g_{k+1}\Delta_k $$
Setting $f_k=s$ and $\Delta g_k = {N-s-1 \choose n-1}$ I remove $s$ from the sum entirely, but this requires guessing an appropriate $g_k$. Is this the right direction to go? 
Ignoring my assumption that summation by parts is possible, are there standard results which can be applied to this sum?

Let us use the identity that $$\sum_{k=0}^{p} {k \choose m}= {p+1 \choose m+1}~~~~(1)$$ Using this let us find
$$\sum_{k=0}^{p} k {k \choose m}=\sum_{k=0}^{p}[ (k+1-1) {k \choose m}= \sum_{k=0}^{p}[(k+1) \frac{k!}{m! (k-m)!}-{k \choose m}]=\sum_{k=0}^{p}[(k+1) \frac{k!}{m! (k-m)!}-{k \choose m}]= \sum_{k=0}^{p}[(m+1) {k+1 \choose m+1}-{k \choose m}]$$
Using )1) we get 
$$\sum_{k=0}^{p} k {k \choose m}= (m+1) {p+2 \choose m+2}-{p+1 \choose m+1}~~~~(2)$$
$$E=\sum_{s=0}^{N-n} s  {N-s-1 \choose n-1} =\sum_{k=n-1}^{N-1} (N-k-1) {k \choose n-1}, ~\text{where}~ N-s-1=k.$$
$$E=\sum_{k=n-1}^{N-1} (N-1) {k \choose n-1}-\sum_{k=n-1}^{N-1} k {k \choose n-1}$$
Using (1) and (2), we get 
$$E=(N-1){N \choose n}-n {N+1 \choose n+1} +{N \choose n}$$
$$E=N {N \choose n}-n {N+1 \choose n+1}$$
Finally $$E={N \choose n+1}$$
as pointed out by @Rob Pratt in the comment below.
