Playing around with a proof of a generalized Hardy-Littlewood Tauberian theorem (I'll state the theorem at the end of this question), I ran into the following proposition:
Let $$F_n(\alpha) = \frac{1}{n}\sum_{k=1}^{n-1} \bigg(\log\frac{n}{k}\bigg)^\alpha.$$ Then $\lim_{n\to\infty} F_n(\alpha) = \Gamma(\alpha+1)$.
I have a proof in the case $\alpha\geq 0$, which is the case I'm interested in. I'll add it as an answer in a moment. But I want to post this as a question because it's interesting, and in case others come up with different proofs and might like to add their answers.
For posterity, here's the generalized Hardy-Littlewood Tauberian theorem. Suppose $\mu$ is a measure on $[0,\infty)$ whose cumulative distribution $D_\mu(x) = \mu([0,x])$ is of bounded variation. Let $\omega(s) = \int_0^\infty e^{-x s}d\mu(x)$ be its Laplace transform. Then: $$D_\mu(x)\sim x^\alpha\ \mbox{ as $x\to\infty$ if and only if }\ \omega(s)\sim s^{-\alpha}\ \mbox{ as $s\to 0$}.$$ I was playing with the proof in Ch 8 of the second volume of Taylor's book Partial Differential Equations.
First, notice that in fact $$F_n(\alpha) = \frac{1}{n}\sum_{k=1}^n \log(n/k)^\alpha,$$ as the last term is equal to zero.
Define $f(t) = -\log(t)^\alpha$. We may reinterpret $F_n$ as a Riemann sum for $f$ over $[0,1]$ for the partition $t_k = k/n$, where $k$ ranges over $1,\ldots,n$. Then $$ F(\alpha) = \lim_{n\to\infty} F_n(\alpha) = \int_0^1 f(t)\ dt = \int_0^1 (-\log t)^\alpha\ dt $$ Now we exploit the fact that the logarithm is monotone: we perform a substitution. Let $s = -\log t$, so that $t = e^{-s}$ and $ds = -dt/t$. Therefore $dt = -e^{-s}ds$. The limits of integration $0$ and $1$ become $\infty$ and $0$, respectively. The minus sign in $ds$ reverses the limits, and voila! $$ F(\alpha) = \int_0^\infty s^\alpha e^{-s}\ ds = \Gamma(\alpha+1).$$