real analysis - For sufficiently large $n$, Which number is bigger, $2^n$ or $n^{1000}$?

How do I determine which number is bigger as $n$ gets sufficiently large, $2^n$ or $n^ {1000}$?

It seems to me it is a limit problem so I tried to tackle it that way.

$$\lim_{n\to \infty} \frac{2^n}{n^{1000}}$$

My thoughts are that, after some $n$, the numerator terms will be more than the terms in the denominator so we'll have something like $$\frac{\overbrace{2\times 2\times\cdots \times 2}^{1000\text{ factors}}}{n\times n \times \cdots \times n} \times 2^{n-1000}$$

At this point, I was thinking of using the fact that $2^n$ grows faster slower than $n!$ as $n$ gets larger so the limit, in this case, will be greater than $1$, meaning $2^n$ is bigger than $n^{1000}$ for sufficiently large $n$. This conclusion is really just a surmise based on a non-concrete formulation. Therefore, I'd appreciate any input on how to tackle this problem.

Answer

Note that $$\frac{2^{n+1}}{2^n}=2\text{ whereas }\frac{(n+1)^{1\,000}}{n^{1\,000}}=\left(1+\frac1n\right)^{1\,000}.$$ Since $$\lim_{n\to\infty}\left(1+\frac1n\right)^{1\,000}=1<2,$$ you have that $$\left(1+\frac1n\right)^{1\,000}<\frac32$$ if $n$ is large enough.It's not hard to deduce from this that $2^n>n^{1\,000}$ if $n$ is large enough.