Prove that $\sqrt [m]{2}$ is irrational for every integer $m\ge2$ So I have a small question about the following proof,
Prove that $\sqrt [m]{2}$ irrational for every integer $m\ge2$
So far this is what I have,
Assume towards contradiction, $\sqrt [m]{2}=p/q$ for $m≥2$. Then $2=\frac{p^m}{q^m}$ so $2q^m=p^m$. Then $p^m$ is even so $p$ is even such that $p=2k$ for some integer, $k$. Then $2q^m=p^m=(2k)^m$. So $q^m=2^{m-1}(k)^m$. 
For the next step I was wondering if I needed a lemma that proves that $2^{m-1}$ is a multiple of $2$ before I can state that $q^m$ is even?
I would appreciate it if anyone could clarify this for me. Thank you!
 A: $2^{m-1}$ is a multiple of $2$ for $m\geq 2$. You don't need to prove this, it follows directly from the rules of exponentiation. 
$2^{m-1}=2 \times 2^{m-2}$ and since $m \geq 2$ we know that $m-2 \geq 0$. For $m=2$ the term $2^{2-2}=1$ but $2 \times 1$ is still a multiple of $2$. The result for $m>2$ follows even easier than this.
A: Assume there is a rational number, a/b, such that (a/b)^m=2. This can be rewritten as (a^m)/(b^m)=2. Multiplying both sides by b^m gives us a^m=2(b^m), or, a^m=b^m+b^m, which is actually Fermat's Last Theorem which has been proven to have no integer solutions. So there are no integers a and b such that (a/b)^m=2, meaning that a/b is irrational.
This proof is taken from a Numberphile video
A: For a prime $\ell$ and an equation $\ell q^m=p^m$ with $m\ge2$, there’s an immediate contradiction to the uniqueness part of the Fundamental Theorem of Arithmetic.
Because: count how many appearances of the prime $\ell$ there are on the two sides of the equation: divisible by $m$ on the right, indivisible by $m$ on the left. So, a contradiction.
