We have given the Diophantine equation $(1) \;\; \frac{k^l}{l^k} = \frac{k!}{l!}$, where $k$ and $l$ are positive integers. Clearly equation (1) is satisfied if $k=l$. Assume $k \neq l$. We may (since $k$ and $l$ are symmetric in equation (1)) WLOG assume $k>l$. If $l \geq 3$, then $k^l < l^k$, i.e. ${\textstyle \frac{k^l}{l^k} < 1 < \frac{k!}{l!}}$, which according to equation (1) is impossible. Consequently $l<3$ by contradiction. Inserting $l=2$ (which implies $k>3$ since $k>l$) in equation (1), the result is $k = 2^{k-1}(k - 1)!$, which has no solution since $k < 2^{k-1}(k - 1)!$ when $k\geq 3$. Finally, by inserting $l=1$ in equation (1), the result is $(k-1)! = 1$, yielding $k=2$. Conclusion: The only solutions of eqaution (1) in positive integers are $(k,l) = (1,2), (2,1), (n.n)$, where $n$ is a positive integer.