I need to show that the variance of a binomial probability distribution Var(X) = npq. You can see a full proof here. I'm working on the $E\left[ { X }^{ 2 } \right] $ term and followed it all until the re-indexing moment, where it looks like $n$ is simply changed to $m$ while it should be that $m=n-1$, so I'd like help with how the adjustment here works. Here's my work, assuming the first few steps and factoring out $np$ are given:
$$np\sum _{ k=1 }^{ n }{ k } \left( \begin{matrix} n-1 \\ k-1 \end{matrix} \right) { p }^{ k-1 }{ (1-p) }^{ n-k } $$
and if we let $m=n-1$ and $j=k-1$,
$$np\sum _{ j=0 }^{ m+1 }{ (j+1) } \left( \begin{matrix} m \\ j \end{matrix} \right) { p }^{ j }{ (1-p) }^{ m-j }$$
I'm stuck with $m+1$ for the upper bound of my index and can't see how to change it to $m$.