I'm trying to follow a proof that any falling power can be converted to a sum of multiples of regular powers, i.e.
$x^{\underline{n} = \sum_{k=0}^n s_{n,k}x^k}$ with $s_{n,n}=1$ and $s_{n,0}=0$ for $n\ge1$.
The proof I'm following is for proposition 1.6.1 in this document, on page 44. The step I don't understand involves simplifying some sums in the inductive step on page 45. I reproduce some of the proof here:
(definition of falling power) $x^{\underline{n}} = x^{\underline{n-1}} \cdot (x-n+1)$
(by inductive hypothesis) $=(x-n+1)\sum_{k=0}^{n-1} s_{n-1,k}x^k $
(I'm still following here) $x\sum_{k=0}^n s_{n-1,k}x^k - (n-1)\sum_{k=0}^{n-1} s_{n-1,k}x^k$
(Huh???) $-(n-1)s_{n-1,0}x^0 + \sum_{k=1}^{n-1}(s_{n-1,k-1} - (n-1)s_{n-1,k})x^k + s_{n-1,n-1}x^n$
(Already lost...) $= 0x^0 + \sum_{k=1}^{n-1}(s_{n-1,k-1}-(n-1)s_{n-1,k})x^k + 1x^n$
I'm lost at the "Huh???" step. The coefficients $s_n,k$ are the Stirling numbers, but the proof just shows that the decomposition is possible. It seems a little superfluous to me to have to prove this in the first place, since $x^{\underline{n}}=x(x-1)(x-2)...$, which is obviously a regular ol' polynomial. However, I would still like to be able to follow the proof, especially since I need to learn to follow proofs with summations better.
Can anyone explain to me the manipulation peformed in the "Huh???" step?
Adding a line before that "huh" line might help. Beginning at the "still following here" line, we have:
\begin{eqnarray*} x^\underline{n} &=& x\sum_{k=0}^{n-1} s_{n-1,k}\,x^k - (n-1)\sum_{k=0}^{n-1} s_{n-1,k}\,x^k \\ && \\ &=& x\sum_{k=1}^n s_{n-1,k-1}\,x^{k-1} - (n-1) s_{n-1,0}\, x^0 - (n-1)\sum_{k=1}^{n-1} s_{n-1,k}\,x^k \\ && \mbox{by re-indexing the first sum and also separating the $k=0$ term from the second sum} \\ && \\ &=& -(n-1) s_{n-1,0}\, x^0 + \sum_{k=1}^{n-1} \left(s_{n-1,k-1} - (n-1) s_{n-1,k}\right)x^k + s_{n-1,n-1}\,x^n \\ && \mbox{the last term is the $k=n$ term in the first sum in the previous line} \\ && \\ &=& 0x^0 + \sum_{k=1}^{n-1} \left(s_{n-1,k-1} - (n-1) s_{n-1,k}\right)x^k + 1x^n \\ && \mbox{by using the inductive hypothesis: $s_{n-1,0} = 0$ and $s_{n-1,n-1}=1$.} \\ && \\ \end{eqnarray*}
The RHS is now in the form the author was seeking, i.e. $\sum_{k=0}^n s_{n,k}\, x^k$.