Djrk-Jlxjd postulate

Since $0$ and $99989$ are two consecutive multiples of $99989$ (which is a prime), we know that $99989\not\mid k~\forall~1\leq k\leq 99988$ where $k\in\Bbb Z$ .

Now, we use Fermat's Little Theorem which states (with prime $p$ ),

$a^{p-1}\equiv 1\pmod{p}~\forall~a\in\Bbb{Z}~\land~p\not\mid a$

The given sum becomes,

$\sum_{k=1}^{99988} k^{99988}\equiv \sum_{k=1}^{99988}1\equiv 99988\pmod{99989}$

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The solution by Gari is correct although I think he should elaborate it to make it more understandable to others.