= Solution
{c}
The <exponential series> gives $x^x=e^{x\log x}=\sum_{k=0}^\infty(x\log x)^k/k!$. Since $|x\log x|\leq1/e$ on $(0,1)$, the <Weierstrass M-test> proves <uniform absolute convergence>, using the summable bounds $e^{-k}/k!$. Therefore one can integrate the <series> term by term over this finite interval. Using part <1/c/i>,
$$
\int_0^1x^x\,dx=\sum_{k=0}^\infty\frac1{k!}\int_0^1(x\log x)^k\,dx
=\boxed{\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^{k+1}}.}
$$
This is the alternating identity in <sophomore's dream>. The same estimate also proves <absolute convergence> of the integrated <series>; no cancellation is needed to justify the interchange.
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