Prove √ is not a rational number

[polatinte]

Homework Statement

"Prove ^m√n is not a rational number for any natural numbers with n,m > 1, where n is not an m^th power"

Homework Equations

Natural numbers for us start at 1.
Since we know n is not an m^th power, then n $\neq$ k^m for an arbitrary integer k.

The Attempt at a Solution

I believe this would follow a similar argument for the classic proof that √2 is irrational.

We set up the contradiction that ^m√n is actually rational. Then a^m=nb^m for some a,b $\in$ $Z$, b $\neq$ 0, a,b are irreducible.
Then a^m is a multiple of n, so a is a multiple of n.

I'm assuming I need to prove this as a lemma.
I've attempted proving this by modular arithmetic.

Prove by contrapositive (If a is not a multiple of n, then a^m is not a multiple of n). But then there would be n-1 cases to check (infinitely many), so I follow with induction.
Base case: n=2, a $\equiv$ 1 (mod 2), then a² $\equiv$ 1² = 1 $\equiv$ 1 (mod 2).
Base case checks out, so we can assume the inductive hypothesis that this is true for all n.
To show show n+1, the (mod n) turns into (mod n+1), so how do I use the inductive hypothesis on this?
This is the point where I get lost. Any suggestions?

 

[jgens]

You can do this proof much more easily just by counting prime factors.