$\sum_{k=0}^n (-1)^k {n \choose k} k^n = (-1)^n n!$ ?!

Question

$$ \sum_{k=0}^n (-1)^k {n \choose k} k^v = \begin{cases} 0 & (v=0,1,\dots,n-1), \\ (-1)^n n! & (v=n). \end{cases} $$

This formula appears in the book "Introduction to Analysis" (written by Teiji Takagi) ~ Chapter 2 "Differential Method" ~ Section 25 "Taylor's Formula" ~ "Additional Notes" (discussing the "difference" in differential calculus). ( Scanned image of that page)

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What I have tried:

$$ \newcommand{\cc}[2]{{\color{#1}#2}} \newcommand{\cR}[1]{\cc{Red}#1} \newcommand{\cO}[1]{\cc{Orange}#1} \newcommand{\cY}[1]{\cc{Yellow}#1} \newcommand{\cG}[1]{\cc{Green}#1} \newcommand{\cB}[1]{\cc{Blue}#1} \newcommand{\cI}[1]{\cc{Indigo}#1} \newcommand{\cV}[1]{\cc{Violet}#1} \begin{align*} \text{LHS} &= -n +\frac{n(n-1)} {2\cdot1} 2^v -\frac{n(n-1)(n-2)} {3\cdot2\cdot1} 3^v + -\cdots +\frac{n(n-1)\cdots1}{n(n-1)\cdots1} n^v \\ & \begin{array}{} = - n \cR( 1 \\ &\kern-1.5em - (n-1) \cO( 2^{v-1} \\ &&\kern-2.8em - (n-2) \cY( \dfrac {3^{v-1}}{2!} \\ &&&\kern-3em - (n-3) \cG( \dfrac {4^{v-1}}{3!} \\ &&&&\kern-3em - (n-4) \cB( \cdots \\ &&&&&\kern-2.2em -\cdots \cI( \cdots \\ &&&&&&\kern-2.2em - 2 \cV( \dfrac{(n-1)^{v-1}}{(n-2)!} \\ &&&&&&&\kern-5.6em - \dfrac {n^{v-1}}{(n-1)!} \cV)\cI)\cB)\cG)\cY)\cO)\cR) \end{array} \end{align*} $$

I'm stuck here, please help me, thanks!

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Side note: For alternating sequences, there is a more advanced question that has been answered.

Answer 1

First step is to solve the $\nu=0,n=0,1,2,3,\dots$ case. From binomial formula: $$ (1+x)^n = \sum_{k=0}^n\binom nk x^k$$ So in particular, $\sum_{k=0}^n\binom nk (-1)^k=(1+(-1))^n = 0.$

The next step is to use a differentiation trick for sums. Observe that $$Tf(x):= x\frac{d}{dx}f(x), \quad T^2 f:=x\frac{d}{dx}\left(x\frac{d}{dx}f(x)\right),\quad \dots $$ satisfies for $\nu\ge 0$ $$ T^\nu x^k=k^\nu x^k.$$ Hence, the sum in question is solved if we can compute the value of $$ T^\nu(1+x)^n = \sum_{k=0}^n \binom nk k^\nu x^k$$ at $x=-1$.

Note that $T1=0$, $T(1+x)=x$, and generally if $n\ge 1$ then $$T(1+x)^n = n(1+x)^{n-1} x = n(1+x)^{n-1} (x+1-1) = n\Big((1+x)^{n} - (1+x)^{n-1}\Big) $$ Note that if $\nu=1=n$, then the $(1+x)^{n-1}=1$ term is not zero even when $x=-1$. But all higher powers of $(1+x)$ at $x=-1$ are zero. If instead $\nu=1<n$, this lowest order term also evaluates to zero: $(1+(-1))^{n-1}=0$. This proves the result for $\nu = 1, n\ge 0$: $$\sum_{k=0}^n \binom nk k (-1)^k = \begin{cases} 0 & n\neq 1 \\ -1 & n=1 \end{cases}$$ Observe the lowest order coefficients of $(1+x)$: \begin{align} T(1+x)^n &= \dots - n(1+x)^{n-1}\\ T^2(1+x)^n &= \dots + n(n-1)(1+x)^{n-2}\\ T^3(1+x)^n &= \dots - n(n-1)(n-2)(1+x)^{n-3}\\ \vdots \quad &\quad \vdots \qquad \vdots \end{align} Inductively, we see that if we write $$ T^\nu (1+x)^n$$ as a sum of powers of $(1+x)$, the lowest order term is $$(-1)^\nu n^{\underline \nu}(1+x)^{n-\nu},$$ where $n^{\underline \nu}=n(n-1)(n-2)\dots(n-\nu+1)$ is the falling power. In particular, $n^{\underline n}=n!$. If $\nu < n$, then this evaluates to $0$, like all the higher order terms. If $\nu=n$, then this evaluates to $(-1)^n n!$; this is exactly what we sought to prove.

Answer 2

It seems Takagi is doing the backward finite difference, see also here. One can feel a bit uneasy about using the letter $\Delta$ for the difference step and for the operator, but it's analogous to the infinitesimals, so it should be fine after a while.

Main point: if $P(x)$ is a polynomial of degree $n$ then $\Delta P(x)$ is a polynomial of degree $n-1$ the the leading coefficient of $\Delta P(x)$ is $n \times$ leading coefficient of $P(x)$.

Now, also check that

$$\Delta^n f(x) = \sum (-1)^k \binom{n}{k} f(x- k h)$$

Now, if $P(x)$ is of degree $<n$ then $\Delta^n P(x) \equiv 0$, while if $P(x)$ is of degree exactly $n$ then $\Delta^n P(x) = n! \times $ leading coefficient of $P(x)$.

The expressions you have are $\Delta^n P(x)$ at the point $n$ for the polynomial $P(x) = x^{\nu}$ ( use also $\binom{n}{k} = \binom{n}{n-k}$), and $h=1$.

Check an example here ( $n=7$, $a$ is $\nu$).