Prove that $\sqrt {2^n-1}$ is irrational for every integer $ n>1$ 
Prove that $\sqrt {2^n-1}$ is irrational for every integer $ n>1$

I tried assuming it was equal to $\frac p q $.
I get $2^nq^2-q^2 = p^2 $ 
But I don't see where to go from there.
 A: For $n\geq2$, $2^n-1\equiv 3\pmod 4$, unlike a perfect square.
A: Landon Carter's answer is direct and has the elegance of simplicity. Here is a partial answer for the case $n=2k$, an even number.
I too use the fact that if the square root is rational it has to be an integer.
So we assume $2^{2k}-1$ is a square and will get a contradiction.
That is, $(2^k-1)(2^k+1)$ is a square. The these bracketed quantities are both odd and differ by 2, hence have no common factors.
So this forces both the numbers i.e., $2^{2k}\pm1$ to be perfect squares.
AT the beginning the squares $1$ and $4$ differ by $3$, afterwards squares have to differ by more than 3.
Here we have two squares differing by 2.
Contradiction.
A: To continue your idea, rather than to restart and do one of the other correct answers.
$2^nq^2 - q^2 = p^2$.
Assume $p = 1$
The $2^nq^2 - q^2 = 1$.  Then $2^n - 1 = 1/q^2$ is an integer.  So $q=1$.  So $2^n - 1 = 1$.  So $2^n = 2$ so $n = 1 \not > 1$.
Assume $p \ne 1$.
Let $k$ be a prime factor of $p$.  Then either $k|q^2$ which isn't possible as we assumed (or should have assumed) $p/q$ is in lowest terms, or $k|2^n - 1$. 
This is true for all prime factors so $p^2|2^n -1$. 
So $q^2\frac{2^n - 1}{p^2} = 1$ and $\frac{2^n - 1}{p^2} \in \mathbb Z$ so $\frac{2^n - 1}{p^2} = 1/q^2 \in \mathbb Z$ so $q = \pm 1$ and
$2^n - 1 = p^2$.
$p$ is clearly odd so let $p = 2p' + 1$
$2^n = 4p'^2 + 4p' + 2$
$4 \not \mid 4p'^2 + 4p +2 = 2^n$ so $n < 2$.  So $n \le 1$.  So $n \not > 1$.
A: It is clearly equivalent to prove that the diophantine equation $$2^n=x^2+1$$ has no integer solution for $n\gt 1$ Notice that $x$ must be odd so $x=2m+1$ hence
$$2^n=(4m^2+4m+1)+1=4m^2+4m+2\iff2^{n-1}=2m^2+2m+1$$ This is absurde.
