Prove the following

$$\sum_{n=k}^\infty(-1)^{n-k}\left[n\atop k\right] \frac{z^n}{n!} = \frac{\log^k(1+z)}{k!}$$

$$\textit{proof}$$

We start by the following

$$(1+z)^u = \sum_{n=0}^\infty {u \choose n} z^n$$

Now use that

$${u \choose n} = \frac{\Gamma(u+1)}{\Gamma(u-n+1)n!}$$

Now use that

$$\frac{\Gamma(u+1)}{\Gamma(u-n+1)} = \frac{u(u-1)\cdots (u-n+1)\Gamma(u+1)}{\Gamma(u+1)} = (u)_n$$

This implies that

$$(1+z)^u = \sum_{n=0}^\infty {u \choose n} z^n = \sum_{n=0}^\infty \frac{z^n}{n!}(u)_n = \sum_{n=0}^\infty \frac{z^n}{n!}\sum_{k=0}^n s(n,k) u^k$$

By interchanging the two sums we have

$$(1+z)^u = \sum_{n=0}^\infty {u \choose n} z^n = \sum_{k=0}^\infty u^k \sum_{n=k}^\infty \frac{z^n}{n!} s(n,k)$$

It is also easy to see that

$$(1+z)^u = \sum_{k=0}^\infty \log^k(1+z) \frac{u^k}{k!}$$

By comparing the two series we have

$$\sum_{n=k}^\infty \frac{z^n}{n!} s(n,k) = \frac{\log^k(1+z)}{k!} $$

Or

$$\sum_{n=k}^\infty (-1)^{n-k}\left[n\atop k\right] \frac{z^n}{n!} = \frac{\log^k(1+z)}{k!} $$

This proof is sketched on Wikipedia.

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