Corollary 1

Let $a$, $a'$, and $b$ be integers such that $a < a'$ and $b > 0$. Then $a^b < {a'}^b$.

We need only express $a^b$ and ${a'}^b$ as products of finite families of integers (§ 3, no. 5, Proposition 10) and apply Proposition 3, observing that the relation $a < a'$ implies $a' > 0$.