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...

Alternatively, $$\mu = \sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom{N}{k}}
= \sum_{m=k}^{N}\frac{m! k!(N-k)!}{N!(k-1)!(m-k)!}$$
$$= k\frac{k! (N-k)!}{N!} \sum_{m=k}^N \binom{m}{m-k}
= k\frac{k! (N-k)!}{N!} \binom{N+1}{N-k}$$
$$= k\frac{k! (N-k)! (N+1)}{N! (N-k)! (k+1)!} = k\frac{N+1}{k+1}$$
where
$$\sum_{m=k}^N \binom{m}{m-k} = \sum_{i=0}^{N-k} \binom{k+i}{i} = \binom{N+1}{N-k}$$
can be seen from Pascal's Triangle.
