Is 'every exponential grows faster than every polynomial?' always true?

My algorithm textbook has a theorem that says

‘For every $r > 1$ and every $d > 0$, we have $n^d = O(r^n)$.’

However, it does not provide proof.

Of course I know exponential grows faster than polynomial in most cases, but is it true for all case?

What if the polynomial function is something like $n^{100^{100}}$ and exponential is $2^n$? Will the latter outgrow the former at some point?

Solutions Collecting From Web of "Is 'every exponential grows faster than every polynomial?' always true?"

Yes, it is true for all cases. This can be seen by noting that

$$\lim_{n\to\infty} \frac{n^k}{e^n} = 0$$

for any $k$. This can be seen by an application of L’Hospital’s rule a number of times, or by using induction as here.

Hint: Yes. See Taylor expansion of exponential function.

Let $n = k^2$.

Then $n^c = k^{2c}$ and $2^n = (2^k)^k$.

Clearly $2^k \ge k+1 > k$ so all we need is $k > 2c$.

In general if we want to find $n \in \mathbb{N}$ such that $r^n > n^c$ where $r > 1$, we can do essentially the same:

Let $n = ak^2$ where $a > \log_r(2)$.

Then $r^n > (2^k)^k$ and $n^c = a^c k^{2c}$.

It is then enough to choose $k$ such that $k > a$ and $k > 3c$, so that $n^c < k^{3c} < (2^k)^k < 2^n$.