Multiply the Powers!

Let $L = m^n \text{ and } R = 2^{mn}$

$\log_2L = \log_2(m^n) \text{ and } \log_2R = \log_2(2^{mn})$

$\log_2L = n\log_2(m) \text{ and } \log_2R = mn\log_2(2)$

$\log_2L = n\log_2(m) \text{ and } \log_2R = mn$

Now, for all $m \in \Z^+$ ,

$\log_2m \lt m$

$n\log_2m \lt nm \ldots (\text{as } n \gt 0)$

$\log_2L \lt \log_2R$

Now, as $\log_2x$ is continuous increasing function, so if $\log_2x_1 \lt \log_2x_2$ , then $x_1 \lt x_2$ . Therefore

$L \lt R$

$\color{#3D99F6}{\boxed{m^n \lt 2^{mn}}}$

Only prime numbers have two factors whose product equals the number itself . Hence $x$ must be a prime number. It has factors $1$ and $x$ itself, such that $m^n=1$ or $x$ , of which $x\geq 2$ is the larger.

$2^{mn}=2^x$ , which is greater than $x$ for all values of $x\geq 2$ .