As part of a larger problem I have a sequence whose limit I am not able to prove. The limit is
$$ \lim_{n\to\infty} \frac{\delta^{2^n-1}}{\gamma^n} \,, $$
where both $0 < \delta < 1$ and $0 < \gamma < 1$ but we do not know the relationship between $\delta$ and $\gamma$ (which is larger). Now numerical experiments indicate that this limit converges to zero even in the extreme case when $\delta$ is just under 1 and $\gamma$ is very small. Clearly this is due to the fact that $2^n-1$ increases much more quickly than $n$. I can't seem to prove this convergence though. Though this evaluates to the indeterminate $0/0$, I shouldn't use L'Hopital. I don't think that helps you anyway since it doesn't get rid of the exponentials. Any ideas here? Am I missing something really simple?
Observe from the Binomial Theorem that $2^{n}-1=(1+1)^n-1\gt \frac{n(n-1)}{2}$ if $n\ge 2$.
So, for $n\ge 2$, our expression is less than $\dfrac{\delta^{(n)(n-1)/2}}{\gamma^n}$.
This is equal to $$\left( \dfrac{\delta^{(n-1)/2}}{\gamma}\right)^n.$$ But $\dfrac{\delta^{(n-1)/2}}{\gamma}$ has limit $0$ as $n\to\infty$, and the result follows.