How prove: $a=x$ and $b=x^x$ for $x^{a+b}=a^b b$? Let $x, a, b$ natural numbers such that $x^{a+b}=a^b b$. How prove: $a=x$ and $b=x^x$? 

Let $p$ be any prime and let $v$ denote the corresponding valuation (i.e., $v(n)=k$ if $p^k\mid n$ and $p^{k+1}\nmid n$).
Then $$\tag1(a+b)v(x)=bv(a)+v(b). $$
If $v(a)<v(x)$, we find $v(b)=av(x)+b(v(x)-v(a))\ge a+b\ge b>v(b)$, contradiction. Therefore $v(a)\ge v(x)$. Say, $v(a)=v(x)+\delta$ with $\delta>0$.
Then $(1)$ becomes
$$ av(a) +(a+b)\delta=v(b)$$
If $\delta>0$, we again arrive at a contradiction:
$$v(b)=av(a)+(a+b)\delta\ge b\delta\ge b>v(b).$$
We conclude $\delta=0$, i.e. $v(a)=v(x)$. As $p$ was arbitrary (and $a,x$ are positive), we conclude $a=x$, and then clearly $b=x^x$.
