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- PROOF OF THE CALCULUS FORMULA
Proof of the identity
$$
\int_0^1\frac{dx}{x^x}=\sum_{n=1}^\infty \frac1{n^n}=\frac1{1^1}+\frac1{2^2}+\frac1{3^3}+\dots\ .
$$
First, recall the integral
$$
\Gamma(n)\equiv\int_0^\infty e^{-s}s^{n-1}\,ds=(n-1)!
$$
Then use the substitution $x=e^{-t}$, the exponential series, shifting the index $n$ by one, the substitution $nt=s$, and $\Gamma(n)=(n-1)!$:
\begin{eqnarray*}
\int_0^1x^{-x}dx &=& \int_0^\infty \left(e^{-t}\right)^{-e^{-t}}e^{-t}\,dt =\int_0^\infty e^{te^{-t}}e^{-t}\,dt\\
&=&\int_0^\infty \left[\sum_{n=0}^\infty \frac1{n!}t^ne^{-nt}\right]e^{-t}\,dt\\
&=&\sum_{n=0}^\infty\frac1{n!}\int_0^\infty t^ne^{-(n+1)t}dt\\
&=&\sum_{n=1}^\infty\frac1{(n-1)!}\int_0^\infty e^{-nt}t^{n-1}\,dt\\
&=&\sum_{n=1}^\infty\frac1{(n-1)!}n^{-n}\int_0^\infty e^{-s}s^{n-1}\,ds\\
&=&\sum_{n=1}^\infty\frac1{(n-1)!}n^{-n}\Gamma(n)= \sum_{n=1}^\infty n^{-n}.
\end{eqnarray*}