In reading about A polarization identity for multilinear maps by Erik G F Thomas, I am led to prove the following combinatorial identity, which I cannot find anywhere, nor do I have any idea how to prove it.
$$\sum\limits_{j=1}^k(-1)^{k-j}j^k\binom{k}{j}=k!.$$ After some reductions (including $\binom{k+1}{j}=\binom{k}{j}+\binom{k}{j-1}$) and "simplifications", I am still stuck and have no clues how to proceed.
I do not know if this is true, either, though I think it should be.

Any help or reference is sincerely welcomed, thanks in advance.


marked as duplicate by Marc van Leeuwen, Community May 23 '16 at 8:15

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  • $\begingroup$ Just noticed it is bionomial theorem for $(-1+j)^k$ with $j\in \mathbb{N}$ $\endgroup$ – José Osorio May 22 '16 at 11:06
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    $\begingroup$ @JoséOsorio But $j$ is not fixed. $\endgroup$ – almagest May 22 '16 at 11:10
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    $\begingroup$ Isn't this the combinatorial formula for the number of surjections from the set of $k$ elements to itself? Proof by Inclusion-Exclusion. $\endgroup$ – lulu May 22 '16 at 11:33
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    $\begingroup$ See also (3) in web.mit.edu/~darij/www/QEDMO6P4long.pdf for an algebraic proof. $\endgroup$ – darij grinberg May 22 '16 at 13:30
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    $\begingroup$ Notice that I prefer the sum to start at $j=0$; otherwise the equality is false for $k=0$. $\endgroup$ – darij grinberg May 22 '16 at 13:32

Let $L$ be the collection of sequences indexed by $\mathbb{N}$ taking values in $\mathbb{R}$. $$L = \{ (x_0, x_1, x_2, \ldots ) : x_i \in \mathbb{R} \}$$ $L$ will be a vector space with respect to componentwise addition and scalar multiplication.
Define a linear map $D$ on $L$ by $$L \ni x = (x_0, x_1, x_2, \ldots ) \quad\mapsto\quad Dx = (x_1, x_2, \ldots ) \in L$$ The result of applying $D - 1$ to a sequence $x$ is the finite difference of $x$. $$\left[ (D-1)x \right]_n = x_{n+1} - x_n$$

If you apply $(D-1)^k$ to the sequence $u = (n^k)_{n=0}^\infty$, you will find

$$\left[ (D-1)^k u \right]_n = \sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j} \left[D^j u\right]_n = \sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j} u_{n+j} = \sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}(j+n)^k $$ When $n = 0$, this is the LHS of the equality at hand.

To see LHS of your equality is indeed equal to the RHS, you can use following fact:

If $v$ is a sequence whose value is a polynomial in the index $n$, i.e. $$v_n = a_p n^p + a_{p-1} n^{p-1} + \cdots + a_0$$ for some constants $a_p, a_{p-1},\ldots, a_0$, the finite difference of $v$ will be a polynomial in $n$ with degree one less. Furthermore, the leading coefficient is multiplied by a factor $p$. i.e. $$\left[(D-1)v\right]_n = p a_p n^{p-1} + \cdots$$

Apply these $k$ times to the sequence $u = (n^k)_{n=0}^\infty$, you find $$\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}(j+n)^k = \left[ (D-1)^k u \right]_n = (D-1)^k n^k = k! n^0 = k!$$ which is a constant independent of $n$ and equal to RHS of the equality at hand.

  • $\begingroup$ Thanks! I have conjectured the last formula as a possible way to prove the special case, but had no idea how to prove it. This helps a lot. :) $\endgroup$ – awllower May 22 '16 at 12:17

A combinatorial answer.

The right-hand side is the number of permutations of $\{0, 1, \dots, k-1\}$.

I'll let $k=5$ for concreteness.

We count the permutations in another way. Take a $5$-tuple drawn from $\{0,1,\dots,4\}$ - that is, a member of $5^5$. This is probably not a permutations - for example, it might be $00000$.

Remove anything which does not contain a $4$. That is, remove anything which can be drawn from $4^5$. This leaves all the permutations, but also leaves things like $44444$, so in fact we should be removing anything which can be drawn from $\{ 1, 2, 3, 4\}$ and from $\{ 0, 2, 3, 4\}$ and so on.

That is, we remove $4^5$ things $\binom{5}{1} = \binom{5}{4}$ times.

OK, we've left all the permutations, but now we've removed some things more than once: we removed $00000$ because it doesn't contain a $4$ but also because it doesn't contain a $3$, for instance. Generally, we've removed twice anything which has two things it doesn't contain, and we've removed three times anything which has three things it doesn't contain, and so on. We need to add them back in enough times: we need to add once anything which has two things it doesn't contain, add twice anything which has three things it doesn't contain, and so on.

Add in $3^5$ things $\binom{5}{2}$ times: namely, add in anything which does not contain a $0$ and does not contain a $1$, then add in anything which does not contain a $0$ and does not contain a $2$, then… then add in anything which does not contain a $3$ and does not contain a $4$. We've re-added anything which has two things it doesn't contain, but we've re-added $\binom{3}{2}$ times anything which has three things it doesn't contain, re-added $\binom{4}{2}$ times anything which has four things it doesn't contain, and so on.

Repeat this procedure of adding and subtracting until we eventually finish.

  • $\begingroup$ Thanks: I was dinning after the post, and my classmate and I just came up with this argument while dining! Quite a coincidence! :D Still thanks for the help! $\endgroup$ – awllower May 22 '16 at 12:16

As lulu said, this can be proved by inclusion-exclusion theorem. We have \begin{equation} |\cup_{i=1}^k A_i|=\sum_{j=1}^k(-1)^{j+1}(\sum_{1\leq i_1<...<i_j\leq k}|A_{i_1}\cap...\cap A_{i_j}|). \end{equation} Now let $A_i$ be the subset of function from $[k]$ to $[k]$ such that $i$ is not an image. Then lhs of the equation is equal to $k^k-k!$, which means we subtrct the bijective functions from all functions. The $j$-th component of rhs is equal to $(-1)^{j+1}\begin{pmatrix}k\\j\end{pmatrix}(k-j)^k$, which means that we choose $j$ elements such that they are not images, and the $k$ elements only have then $k-j$ possibilities to choose an image. Now let $\tilde{j}=k-j$, and inserting this into the equation, you will find the solution of the given form.

More details: we obtain then \begin{align} k^k-k!&=\sum_{j=0}^{k-1}(-1)^{k-j+1}\begin{pmatrix}k\\k-j\end{pmatrix}j^k\\ &=-1\sum_{j=0}^{k-1}(-1)^{k-j}\begin{pmatrix}k\\j\end{pmatrix}j^k\\ &=-1\sum_{j=1}^{k}(-1)^{k-j}\begin{pmatrix}k\\j\end{pmatrix}j^k+(-1)^{k-k}\begin{pmatrix}k\\k\end{pmatrix}k^k\\ &=-1\sum_{j=1}^{k}(-1)^{k-j}\begin{pmatrix}k\\j\end{pmatrix}j^k+k^k. \end{align} Both sides subtrct $k^k$ and devided by $-1$, we obtain the result.

  • $\begingroup$ I am confused: where is the $k!$ factor? Also, I think $(-1)^{j+1}$ becomes $(-1)^{k-\overline{j}+1},$ but the factor in the equation in the question is $(-1)^{k-j} ?$ $\endgroup$ – awllower May 22 '16 at 12:28
  • $\begingroup$ @awllower notice your index of $\tilde{j}$ only varies form $0$ to $k-1$, and in your formula the index is from $1$ to $k$, so there are some more terms appear, and they will give you $k!$. $\endgroup$ – Peter May 22 '16 at 12:30
  • $\begingroup$ @awllower for the $-1$ term, you have the identity $(-1)^{k-j+1}=(-1)(-1)^{k-j}$. $\endgroup$ – Peter May 22 '16 at 12:31
  • $\begingroup$ ok let me add the details in the answer :) $\endgroup$ – Peter May 22 '16 at 12:33
  • $\begingroup$ @awllower I think I also made an error, the lhs is then $k^k-k!$, but not the former one, sorry for the confusion. $\endgroup$ – Peter May 22 '16 at 12:50

In this answer, it is shown that for any $l$, $$ \begin{align} \sum_{r=0}^n\binom{n}{r}(-1)^r(l-r)^n &=\sum_{r=0}^n(-1)^{n-r}\binom{n}{r}(r-l)^n\\ &=n! \end{align} $$ The given identity is this identity using $l=0$.

  • $\begingroup$ Thanks for referencing the question, where many nice answers are! $\endgroup$ – awllower May 22 '16 at 14:06

Here is a variant based upon the coefficient of operator $[z^k]$ used to denote the coefficient of $z^k$ in a series. This way we can write e.g. \begin{align*} [z^j](1+z)^k=\binom{k}{j}\qquad\qquad\text{or}\qquad\qquad k![z^k]e^{jz}=j^k \end{align*}

We obtain \begin{align*} \sum_{j=1}^k(-1)^{k-j}j^k\binom{k}{j} &=\sum_{j=0}^\infty(-1)^{k-j}k![z^k]e^{jz}[u^k](1+u)^k\tag{1}\\ &=(-1)^kk![z^k]\sum_{j=0}^{\infty}\left(-e^{z}\right)^j[u^k](1+u)^k\tag{2}\\ &=(-1)^kk![z^k](1-e^z)^k\tag{3}\\ &=k![z^k]\left(z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right)^k\tag{4}\\ &=k! \end{align*} and the claim follows.


  • In (1) we apply the coefficient of operator twice and extend the lower and upper limit of the series without changing anything, since we add only zeros.

  • In (2) we use the linearity of the coefficient of operator and do some rearrangements.

  • In (3) we use the substitution rule with $u=-e^z$:

\begin{align*} A(z)=\sum_{j=0}^\infty a_jz^j=\sum_{j=0}^\infty z^j [u^k]A(u) \end{align*}

  • In (4) we see that in each of the $k$ factors $(e^z-1)$ only $z^1$ provide a contribution to $z^k$.
  • $\begingroup$ Thanks for this interesting use of the coefficient operator which makes the approach through exponential functions more natural. :) $\endgroup$ – awllower May 23 '16 at 8:16
  • $\begingroup$ @awllower: You're welcome! :-) $\endgroup$ – Markus Scheuer May 23 '16 at 8:28

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