= Solution
Substitute $x=e^{-t}$ to get
$$
\int_0^1(x\log x)^k\,dx=(-1)^k\int_0^\infty t^ke^{-(k+1)t}\,dt
=\boxed{\frac{(-1)^kk!}{(k+1)^{k+1}}},
$$
where the second substitution $u=(k+1)t$ reduces the integral to the supplied gamma integral.
On $(0,1)$, $-x\log x\geq0$. Expand $x^{-x}=e^{-x\log x}=\sum_{j=0}^\infty(-x\log x)^j/j!$. The partial sums increase, so the <monotone convergence theorem> permits termwise integration. The integral of the $j$-th term is $1/(j+1)^{j+1}$, including $j=0$. Reindexing gives
$$
\boxed{\int_0^1x^{-x}\,dx=\sum_{k=1}^\infty\frac1{k^k}.}
$$
This is finite, for instance because the summands with $k\geq2$ are bounded by $2^{-k}$.
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