Problem 27 states: If $S_n(x)=x^n$, and $0\le k\le n$, prove that: $$S_n^{(k)}(x)=\frac{n!}{(n-k)!}x^{n-k}$$
The answer book says:
Proof by induction on k. The result is true for $k=0$. If $$S^{(k)}(x)=\frac{n!}{(n-k)!}x^{n-k}$$
then $$\quad\quad\quad S_n^{(k+1)}(x)=\frac{n!(n-k)}{(n-k)!}x^{n-k-1}\quad\quad\require{enclose}\enclose{circle}{1}$$ $$\quad\quad\quad\quad\quad\quad\,\, =\frac{n!}{[n-(k+1)]!}x^{n-(k+1)}$$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I feel some process is missing in $\require{enclose}\enclose{circle}{1}$. I know it's true, because previous exercise asked to prove: $$ if:\quad S_m(x)=x^m, \quad\quad then:\quad S^{'}_m(x)=mx^{m-1}$$ Let $m=n-k$, is it sufficient to say "Rising $x$ to $-1$ power will multiply $n-k$ on its left." for $\mathbf {any\,\; higher-order}$ derivative?
Another question, should the induction require the result to be true for $k=1$,but not $k=0$?
I know the answer is right, but I missed something here. Thank you!
The idea is not to raise $x$ to $-1$. If you differentiate $x^{n-k}$, what you get is $(n-k)x^{n-k-1}$. That's all that is needed (and all that is used).
And the induction proof starts by proving that the statement holds when $k=0$. The other step is to prove that if it holds for a cetain $k$, it also holds for $k+1$.