Expand $x^x=e^{x\log x}={\sum }_{j=0}^{\infty }(x\log x{)}^j/j!$. The sum of the integrals of the absolute values of its terms is ${\sum }_{k=1}^{\infty }{k}^{-k}<\infty$ by part (c). Absolute summability therefore permits interchange of the series and the integral. Applying the moment formula from part (c),Equivalently, the absolute exponential series is $x^{-x}$, an integrable majorant, so dominated convergence gives the same interchange.

Substitute $x=e^{-t}$ to getwhere the second substitution $u=(k+1)t$ reduces the integral to the supplied gamma integral.

On $(0,1)$, $-x\log x\ge 0$. 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 givesThis is finite, for instance because the summands with $k\ge 2$ are bounded by $2^{-k}$.

The partial sums $S_n={\sum }_{k=1}^n{a}_k$ are nonnegative measurable functions and increase pointwise to the extended-valued sum $S={\sum }_{k=1}^{\infty }{a}_k$. The monotone convergence theorem and finite linearity giveA nonnegative measurable function with finite integral is finite almost everywhere; if it were infinite on a positive-measure set, its integral would be infinite. Define it arbitrarily on the null exceptional set if a real-valued representative is desired. Hence the series converges almost everywhere to an integrable function and

The monotone convergence theorem states that for measurable functions $0\le f_1\le f_2\le \cdots$ converging pointwise to $f$, the limit is measurable andBoth sides may be infinite. Monotonicity and nonnegativity replace any need for an integrable dominating function.

Given a prime factorization $n=p_1\cdots p_r$, repeatedly apply part (c) to the cyclic group of order $n$: it has a chain of subgroups of successive ordersRead in increasing order, this is a composition series with factors isomorphic to the cyclic groups of orders $p_1,\dots ,p_r$. Any second prime factorization of the same integer produces another composition series of the same group $C_n$.

The Jordan–Hölder theorem equates the two multisets of simple factors. Since $C_p\cong C_q$ for primes $p,q$ implies $p=q$ by equality of group orders, the two prime multisets agree. HenceFor $n=1$ both factor lists are empty, corresponding to the trivial group.