N less than M to the N

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Theorem

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

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

$\blacksquare$

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