How to prove that $$ \sum_{k=0}^n(-1)^{n-k}{n \choose k} k^n = n! $$ using mathematical induction? Please, do not use definition of Stirling number etc. algebra tricks.


Suggested intro: maybe these solutions don't meet the OP's specific requirements, but I think that they're worth making available anyway.

$\color{red}{n!}$ is the number of bijective functions from $A=\{1,2,\ldots,n\}$ to $A$.

Let we say that a function $f:A\to A$ has type $k$ if $|f(A)|=k$. The number of functions with type $\leq n$ is simply given by $n^n$, i.e. the number of all possible functions from $A$ to $A$. The number of functions with type $\leq(n-1)$ is given by $\binom{n}{n-1}$ (the number of ways for choosing $(n-1)$ elements in $A$) times $(n-1)^n$ (the number of functions from $A$ to a set with $(n-1)$ elements).
By the inclusion-exclusion principle, the number of functions with type $n$ (i.e. the number of bijective functions) is given by:

$$ n^n-\binom{n}{n-1}(n-1)^n+\binom{n}{n-1}(n-2)^n-\ldots = \color{red}{\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k^n}. $$

Alternative proof. Let $\delta$ be the operator that maps a polynomial $p(x)$ into $p(x+1)-p(x)$.
The following properties are trivial to prove:

  1. If $\partial p$ (the degree of $p$) is $\geq 1$, the degree of $\delta p$ is $\partial p-1$;
  2. If the leading term of $p(x)$ is $a x^n$, the leading term of $\delta p(x)$ is $an x^{n-1}$, i.e. $\delta$ acts on the leading term like the derivative $\frac{d}{dx}$.

Since our sum is just $\delta^n p(0)$ with $p(x)=x^n$, by (2.) it follows that our sum equals $n(n-1)(n-2)\cdots 1 = n!$.

  • $\begingroup$ the OP did ask for a solution by induction without tricks :) $\endgroup$ – gt6989b Sep 26 '16 at 16:47
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    $\begingroup$ @gt6989b: What is the point of doing math avoiding tricks? Tricks are the heart of problem solving. $\endgroup$ – Jack D'Aurizio Sep 26 '16 at 16:49
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    $\begingroup$ Please don’t delete it; both solutions are worth having around, and (as usual) I’m partial to the first one. What you might do is preface it with a statement to the effect that you realize that these solutions don’t meet the OP’s specific requirements, but you think that they’re worth making available anyway; I’ve done that on occasion. $\endgroup$ – Brian M. Scott Sep 26 '16 at 19:39
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    $\begingroup$ I like the second proof. It's a bit like what happens with the binomial transform, and the form of the sum suggests making use of it, however it does not work because $k^n$ depends on $n$. $\endgroup$ – Jean-Claude Arbaut Sep 27 '16 at 2:07
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    $\begingroup$ +1. Nice answer. I'm expecting sometimes in the future a question with the condition "don't write". $\endgroup$ – Felix Marin Sep 27 '16 at 6:52

Actually carrying out a proof by induction seems to be quite difficult. I found it easiest to prove the following stronger result (by induction on $n$):

$$\sum_{k=0}^n(-1)^{n-k}\binom{n}kk^m=\begin{cases} 0,&\text{if }0\le m<n\\ n!,&\text{ if }m=n\;. \end{cases}\tag{1}$$

Assume that $(1)$ holds for a specific $n$. Then

$$\begin{align*} (n+1)!&=(n+1)\sum_{k=0}^n(-1)^{n-k}\binom{n}kk^n\\ &=\sum_{k=0}^n(-1)^{n-k}(n+1-k)\binom{n+1}kk^n\\ &=\sum_{k=0}^n(-1)^{n-k}(n+1)\binom{n+1}kk^n-\sum_{k=0}^n(-1)^{n-k}\binom{n+1}kk^{n+1}\\ &=\sum_{k=0}^n(-1)^{n+1-k}\binom{n+1}kk^{n+1}\\ &\qquad-\sum_{k=0}^n(-1)^{n+1-k}(n+1)\binom{n+1}kk^n\\ &=\sum_{k=0}^{n+1}(-1)^{n+1-k}\binom{n+1}kk^{n+1}-(n+1)^{n+1}\\ &\qquad-(n+1)\sum_{k=0}^n(-1)^{n+1-k}\binom{n+1}kk^n\\ &=\sum_{k=0}^{n+1}(-1)^{n+1-k}\binom{n+1}kk^{n+1}-(n+1)^{n+1}\\ &\qquad+(n+1)\sum_{k=0}^n(-1)^{n-k}\left(\binom{n}k+\binom{n}{k-1}\right)k^n\;. \end{align*}$$

We’d like to show that


so we’d like to show that


or, equivalently, that


And in fact

$$\begin{align*} \sum_{k=0}^{n+1}(-1)^{n+1-k}\binom{n+1}kk^n&=\sum_{k=0}^{n+1}(-1)^{n+1-k}\left(\binom{n}k+\binom{n}{k-1}\right)k^n\\ &=\sum_{k=0}^{n+1}(-1)^{n+1-k}\binom{n}kk^n+\sum_{k=0}^{n+1}(-1)^{n+1-k}\binom{n}{k-1}k^n\\ &=-\sum_{k=0}^{n+1}(-1)^{n-k}\binom{n}kk^n+\sum_{k=0}^n(-1)^{n-k}\binom{n}k(k+1)^n\\ &=\sum_{k=0}^n(-1)^{n-k}\binom{n}k\big((k+1)^n-k^n\big)\\ &=\sum_{k=0}^n(-1)^{n-k}\binom{n}k\sum_{m=0}^{n-1}\binom{n}mk^m\\ &=\sum_{m=0}^{n-1}\binom{n}m\sum_{k=0}^n(-1)^{n-k}\binom{n}kk^m\\ &=0\;, \end{align*}$$

since by the induction hypothesis


for $0\le m<n$. This shows that


and it only remains to show that


when $0\le m\le n$.


$$\begin{align*} \sum_{k=0}^{n+1}(-1)^{n+1-k}\binom{n+1}kk^m&=\sum_{k=0}^{n+1}(-1)^{n+1-k}\left(\binom{n}k+\binom{n}{k-1}\right)k^m\\ &=\sum_{k=0}^n(-1)^{n+1-k}\binom{n}kk^m+\sum_{k=0}^{n+1}(-1)^{n+1-k}\binom{n}{k-1}k^m\\ &=-\sum_{k=0}^n(-1)^{n-k}\binom{n}kk^m+\sum_{k=0}^n(-1)^{n-k}\binom{n}k(k+1)^m\;. \end{align*}$$

If $m=n$, this is

$$\begin{align*} -n!+\sum_{k=0}^n(-1)^{n-k}\binom{n}k(k+1)^n&=-n!+\sum_{k=0}^n(-1)^{n-k}\binom{n}k\sum_{\ell=0}^n\binom{n}\ell k^\ell\\ &=-n!+\sum_{\ell=0}^n\binom{n}\ell\sum_{k=0}^n(-1)^{n-k}\binom{n}kk^\ell\\ &=-n!+\sum_{k=0}^n(-1)^{n-k}\binom{n}kk^n+\sum_{\ell=0}^{n-1}\binom{n}\ell\sum_{k=0}^n(-1)^{n-k}\binom{n}kk^\ell\\ &=-n!+n!+\sum_{\ell=0}^{n-1}\binom{n}\ell\sum_{k=0}^n(-1)^{n-k}\binom{n}kk^\ell\\ &=\sum_{\ell=0}^{n-1}\binom{n}\ell\cdot0\\ &=0\;. \end{align*}$$

And if $0\le m<n$, it’s simply

$$\begin{align*} \sum_{k=0}^n(-1)^{n-k}\binom{n}k(k+1)^m&=\sum_{k=0}^n(-1)^{n-k}\binom{n}k\sum_{\ell=0}^m\binom{m}\ell k^m\\ &=\sum_{\ell=0}^m\binom{m}\ell\sum_{k=0}^n(-1)^{n-k}\binom{n}kk^m\\ &=\sum_{\ell=0}^m\binom{m}\ell\cdot0\\ &=0. \end{align*}$$

This establishes $(2)$ and concludes the induction step.



Base Case $n=1$, check whether the equality holds (you should have just 2 terms in the sum)

Inductive Step

Assume this holds for $n = 1, 2, \ldots, N$ and show this for $n = N+1$. You will have to simplify the terms of the summation to express $$ \sum_{k=0}^{N+1} (-1)^{N+1-k}{N+1 \choose k} k^{N+1} $$ in terms of $$ \sum_{k=0}^{N} (-1)^{N-k}{N \choose k} k^{N} $$


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