N less than M to the N

[edit] Theorem

$\forall m, n \in \Z_+: m > 1 \implies n < m^n$

$n$	$=$	$1 + 1 + \cdots \left({n}\right) \cdots + 1$
	$<$	$1 + m + m^2 + \cdots + m^{n + 1}$	(as $m > 1$ )
	$=$	$\frac {m^n - 1} {m - 1}$	Sum of Geometric Progression
	$\le$	$m n - 1$	(as $m-1 \ge 1$ )
	$<$	$m n$

$\blacksquare$