calculus - Limit of sequence, squeeze theorem?

I have this question, Find the limit of the sequence $$a_n:= \frac{n^{2001}}{1.001^n}$$ as $n \to \infty$.

I presume that the limit is $0$ due to the exponential in the denominator, and also presume I am to use the squeeze theorem to show this, but I am finding it hard to find two bounds that tend to the same limit. Or do I need to use a different theorem?

We have not used logarithms to solve limits yet and this exercise is meant to be completed using theorems and rules such as squeeze theorem, ratio test, sum/product/quotient rules etc.

Answer

Ok, so ratio it is:

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{2001}}{1.001^{n+1}}\cdot\frac{1.001^n}{n^{2001}}=\left(1+\frac1n\right)^{2001}\cdot\frac1{1.001}\xrightarrow[n\to\infty]{}\frac1{1.001}<1$$

and thus the sequence converges to zero.