TITLE: Prime divisibility concerning primes of the form $4k+1$
QUESTION [5 upvotes]: Let $p=4k+1$ be a prime number.
I need to  prove that $p$ divides $k^k-1$.
Could anyone give me some hint?
Thank

REPLY [5 votes]: Note,  mod $\,p = 4k\!+\!1,\ $ that $\,-1\,$ is a square, say $\, -1 \equiv i^2,\ $ by $ $ Euler's criterion. $\phantom{I_{I_{I_I}}}$
$\!\begin{eqnarray}
\text{Therefore}\quad\ {-4}k\equiv 1\,\Rightarrow\,\color{#c00}{k^{-1}\ \equiv\ {-}4}&&\equiv \color{#c00}{(1+i)^4}\ \  {\rm by\ squaring}\ \ 2i \equiv (1+i)^2 \\[0.1em]
\ {\rm Thus\ prior}^k\Rightarrow\, k^{-k}\equiv (\color{#c00}{k^{-1}})^{k} \equiv (\color{#c00}{-4})^{k}\!\!\! &&\equiv \color{#c00}{(1+i)}^{\color{#c00}4k}\!\equiv (1+i)^{p-1}\!\equiv 1\, \Rightarrow\, k^k \equiv 1^{-1}\!\equiv 1.\ \ \, {\rm QED}
\end{eqnarray}$