Let $S_n(u) = \sum_{k=1}^{n}k^u$. Show that the value of $\sum_{u=1}^{m}$ ${m+1 \choose u}S_n(u)$is $(1+n)^{m+1} - (1+n)$.
What I have tried so far is this:
$ \sum_{u=1}^{m}{m+1 \choose u}S_n(u)= \sum_{k=1}^{n} \sum_{u=1}^{m}{m+1 \choose u}k^u$, and from here I guess you have to use the binomial expansion, but I don't know how exactly as then I can't evaluate the other sum.
$$ \begin{aligned} \sum_{u=1}^m{{m+1}\choose{u}}S_n(u) &= \sum_{u=1}^m{{m+1}\choose{u}}\sum_{k=1}^nk^u = \\ &= \sum_{k=1}^n\left(\sum_{u=1}^{m}{{m+1}\choose{u}}k^u\right) = \\ &= \left[ \begin{aligned} (k+1)^{m+1} &= \sum_{u=0}^{m+1}{{m+1}\choose{u}}k^u = 1 + \sum_{u=1}^{m}{{m+1}\choose{u}}k^u + k^{m+1} \Leftrightarrow \\ \sum_{u=1}^{m}{{m+1}\choose{u}}k^u &= (k+1)^{m+1} - 1-k^{m+1} \end{aligned} \right] = \\ &= \sum_{k=1}^n\left((k+1)^{m+1} - 1-k^{m+1}\right) = \sum_{k=1}^n\left((k+1)^{m+1} - k^{m+1}\right) - \underbrace{\sum_{k=1}^n1}_{n} = \\ &= \underbrace{\left(2^{m+1}-1^{m+1}\right) + \left(3^{m+1}-2^{m+1}\right) + \left(4^{m+1}-3^{m+1}\right) + \ldots + \left((n+1)^{m+1}-n^{m+1}\right)}_{\text{telescopic sum} = -1^{m+1} + (n+1)^{m+1} = (n+1)^{m+1}-1}-n = \\ &= (n+1)^{m+1} - 1 - n = \\ &= (n+1)^{m+1} - (1 + n) \end{aligned} $$