Is it true that $2^n$ is $O(n!)$? I had a similar problem to this saying:
Is it true that $n!$ is $O(2^n)$?
I got that to be false because if we look at the dominant power of $n!$ it results in $n^n$. So because the base numbers are not the same it is false.
Is it true that $2^n$ is $O(n!)$?
So likewise with the bases, this question should result in false, however it is true. Why? Is the approach I am taking to solve these questions wrong?

Another approach:
Consider: $$\sum_{n=1}^{\infty}\frac{2^n}{n!}\tag{1}$$
We have that $\limsup_{n\rightarrow \infty}\sqrt[n]{\frac{2^n}{n!}}=\limsup_{n\rightarrow \infty}\frac{2}{\sqrt[n]{n!}}=0$
Because 
$$lim_{n\rightarrow \infty}\sqrt[n]{n!}=lim_{n\rightarrow \infty}(n\cdot (n-1)\cdot (n-2)\dots2\cdot 1)^{\frac{1}{n}}\geq lim_{n\rightarrow \infty} (\frac{n}{2}\cdot \frac{n}{2}\dots \frac{n}{2})^{\frac{1}{n}}\geq lim_{n\rightarrow \infty}(\frac{n}{2})^{\lfloor \frac{n}{2}\rfloor\cdot \frac{1}{n}}=\infty$$
Hence the series in (1) is convergent by the root test. So $\frac{2^n}{n!}$ is a zero-sequence from which the statement follows.
Note that you can replace $2$ by any other number.
