Corollary 2

Let $a$, $b$, and $b'$ be integers such that $a > 1$ and $b < b'$; then $a^b < a^{b'}$.

For there exists an integer $c > 0$ such that $b' = b + c$ (Proposition 2); since $c \geqslant 1$, we have $a^c \geqslant a > 1$, whence $a^{b'} = a^b a^c > a^b$.