If $a\mid b^2, b^2\mid a^3,\ldots ,a^n\mid b^{n+1},b^{n+1}\mid a^{n+2},\ldots$ then $a=b$ I'm stuck with this problem :
Let $a,b$ positive integers such that
$$a\mid b^2, b^2\mid a^3,\ldots ,a^n\mid b^{n+1},b^{n+1}\mid a^{n+2},\ldots$$
Show that $a=b$.
If were $ b > a $ then $\lim_{n \to \infty}\frac{b^n}{a^n}=0 $ choosing $\epsilon = \frac{1}{a}$ we got a contradiction, but I can't show that $b<a$ can't hold.
Any help is apreciated.

Let $p$ be a prime. Assume $p^k\|a$ (i.e., $p^k\mid a$ and $p^{k+1}\nmid a$) and $p^l\|b$. Then from $a^n\mid b^{n+1}$ we get $nk\le (n+1)l$ and from $b^{n+1}\mid a^{n+2}$ we get $(n+1)l\le (n+2)k$. Hence 
$$ -k\le (n+1)(l-k)\le k$$
or
$$ (n+1)|l-k|\le k.$$
For $n\gg k$ we obtain $|l-k|<1$ hene $l=k$. As this holds for all primes $p$ and $a,b$ are positive, we conclude $a=b$.
