Why does this expected value simplify as shown? I was reading about the german tank problem and they say that in a sample of size $k$, from a population of integers from $1,\ldots,N$ the probability that the sample maximum equals $m$ is:
$$\frac{\binom{m-1}{k-1}}{\binom{N}{k}}$$
This make sense. But then they take expected value of the sample maximum and claim:
$$\mu = \sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom{N}{k}} = \frac{k(N+1)}{k+1}$$
And I don't quite see how to simplify that summation. I can pull out the denominator and a $(k-1)!$ term out and get:
$$\mu = \frac{(k-1)!}{\binom{N}{k}} \sum_{m=k}^N m(m-1) \ldots (m-k+1)$$
But I get stuck there...

We know that $$\mu = \sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom{N}{k}}$$
$$= \sum_{m=k}^{N}m \frac{(m-1)! k!(N-k)!}{N!(k-1)!(m-k)!}$$
$$= \frac{k!(N-k)!}{N!(k-1)!} \sum_{m=k}^{N} m \frac{(m-1)!}{(m-k)!}$$
$$ = \frac{k (N-k)!}{N!} \sum_{m=k}^{N} \frac{m!}{(m-k)!}$$
The expression on the right is a sum of falling factorials. 
