The following identity is true for $n\geq1$:

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

You can obtain it from the equation in this question by setting the variables equal to 1.

I was wondering if anyone could come up with an elementary proof, maybe a counting argument? (I've found this rather tricky)

  • $\begingroup$ Turns out this is a duplicate of this question, sorry $\endgroup$ – Blunka Feb 17 '15 at 10:16

There are a couple of proofs here.

It really looks like inclusion-exclusion here, so I would go for the accepted answer in the post I linked to (the other one, however, also appeals to inclusion-exclusion), but at the moment I do not understand fully the argument there. Will try and come back later.

  • 1
    $\begingroup$ Nice find. I honestly did do some searching. $\endgroup$ – Blunka Feb 17 '15 at 10:15
  • $\begingroup$ Thanks. It seems the Stirling argument is the right one, as it provides your counting argument as well. $\endgroup$ – Andreas Caranti Feb 17 '15 at 10:16

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