Question

Prove that if $n$ and $r$ are positive integers
$n^r-n{(n-1)}^r+\frac{n(n-1)}{2!}{(n-2)}^r-\frac{n(n-1)(n-2)}{3!}{(n-3)}^r+\cdots$
is equal to $0$ if $r$ be less than $n$, and to $n!$ if $r=n$.

Answer 1

We have

${(e^x-1)}^n={(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots )}^n$
$=x^n+$ terms containing higher powers of $x$ ..... $\left(1\right)$
Again, by the Binomial Theorem,
${(e^x-1)}^n=e^{nx}-n{e}^{(n-1)x}+\frac{n(n-1)}{1\cdot 2}{e}^{(n-2)x}-\cdots$ ......(2).
By expanding each of the terms $e^{nx},e^{(n-1)x},\dots$ we find that the coefficient of $x^r$ in (2) is
$\frac{n^r}{r!}-n\cdot \frac{{(n-1)}^r}{r!}+\frac{n(n-1)}{2!}\cdot \frac{{(n-2)}^r}{r!}-\frac{n(n-1)(n-2)}{3!}\cdot \frac{{(n-3)}^r}{r!}+\cdots$
and by equating the coefficients of $x^r$ in (1) and (2) the result follows.