I'm aware of proofs for this question such as using Kirchoffs matrix theorem or using a bijection to get $N^{N-2}$ however my first instinct was to use the idea there are ${N\choose 2}$ possible edges in $k_n$ and that any tree with $N$ vertices has $N-1$ edges therefore I tried to simplify:
${{N\choose 2}\choose N-1} = {(N(N-1))/2}!/{(N-1)!}$ where I believe the denominator gets eliminated after $(N-1)^2$ iterations of the numerator.
This is as far as I get with this though, I'm quite new to combinatorics and factorials in terms of simplifying and manipulating equations. The main question is can I get to final answer this way and how would I simply this further?