Ordering of real numbers compatible with n-th powers/reciprocal powers (induction)

Question

I have to use induction to prove that $$0 \leq a < b \implies 0 \leq a^n < b^n$$ for all natural n. Also (perhaps very similarly) that $$0 \leq a < b \implies 0 \leq a^{1/n} < b^{1/n}.$$

I have a working proof for the first one, but it's very long-winded, and involves considering the possible cases for $a = 0$ and $a > 0$ with each of the inequalities in my induction step (so four separate arguments).

I'm wondering if there is some neater way to prove this (still inductively), perhaps using some of the ordering axioms?

Thanks.

Answer 1

Hint: $a^{n+1} = a \cdot a^n < a \cdot b^n < b \cdot b^n = b^{n+1}$.