  Let $\tau(n)$ be the number of divisors of $n$. We give an elementary proof
of the fact that $$ \sum_{n\le x} \tau(n)^r =xC_{r} (\log x)^{2^r-1}+O(x(\log
x)^{2^r-2}), $$ for any integer $r\ge 2$. Here, $$ C_{r}=\frac{1}{(2^r-1)!}
\prod_{p\ge 2}\left( \left(1-\frac{1}{p}\right)^{2^r} \left(\sum_{\alpha\ge 0}
\frac{(\alpha+1)^r}{p^{\alpha}}\right)\right). $$
