# N less than M to the N

## Theorem

$\forall m, n \in \Z_{>0}: m > 1 \implies n < m^n$

## Proof

 $\ds n$ $=$ $\ds \underbrace {1 + 1 + \cdots + 1}_{\text {n times} }$ $\ds$ $<$ $\ds 1 + m + m^2 + \cdots + m^{n - 1}$ as $m > 1$ $\ds$ $=$ $\ds \frac {m^n - 1} {m - 1}$ Sum of Geometric Sequence $\ds$ $\le$ $\ds m^n - 1$ as $m - 1 \ge 1$ $\ds$ $<$ $\ds m^n$

$\blacksquare$