Proving $\lim\limits_{n \to \infty} \frac{n^a}{c^n} = 0$ using L'Hôpital's Rule I am trying to prove $\displaystyle \lim_{n \to \infty} \frac{n^a}{c^n} = 0$ using L'Hôpital's Rule, but I'm stuck.
Here's what I have so far:
$$ \lim_{n \to \infty} \frac{n^a}{c^n} = \lim_{n \to \infty}\frac{an^{n-1}}{c^n \ln c} = \lim_{n \to \infty}\frac{a(a-1)n^{a-2}}{c^n(\ln c)^2 + c^n \frac{1}{c}}$$
All three limits above seem to evaluate to $\frac{\infty}{\infty}$, so I feel like I'm not getting anywhere. Any ideas?

Edit: So, with the help of the hints below, I was able to figure out that
$$ \lim_{n \to \infty} \frac{n^a}{c^n} = \frac{a}{\ln c} \cdot \lim_{n \to \infty} \frac{n^{a-1}}{c^n} = \frac{a}{\ln c} \cdot \frac{a - 1}{\ln c} \cdot \lim_{n \to \infty} \frac{n^{a-2}}{c^n} = \cdots $$
So, disregarding the constant, it looks like the numerator keeps decreasing, while the denominator stays the same.
I can also see that if I let $a = 2$, for instance, I end up with $0$ after applying L'Hopital's $2$ times:
$$ \begin{aligned} \lim_{n \to \infty} \frac{n^2}{c^n} &\overset{LH}= \lim_{n \to \infty} \frac{2n}{c^n \ln c} \\ &= \frac{2}{\ln c} \lim_{n \to \infty} \frac{n}{c^n} \\&\overset{LH}= \frac{2}{\ln c} \lim_{n \to \infty} \frac{1}{c^n \ln c} \\ &= \frac{2}{(\ln c)^2} \lim_{n \to \infty} \frac{1}{c^n} \\ &= 0 \end{aligned} $$
So it seems reasonable to conclude that for an arbitrary $a > 0$, I will end up with $0$ after applying L'Hopital's $a$ times.
But I'm not sure how to go about using induction to prove it formally. I've only proven very simple sums by induction so far. Do I have to apply it to a product here?


Hint If $c>1$, the limit is trivial for $a \leq 0$. For $a>0$ show instead that 
$$\left( \lim\limits_{n \to \infty} \frac{n^a}{c^n}  \right)^\frac{1}{a}=0$$
Then, raise both powers to $a$.
