You can also use the fact that , well-known as Stirling's approximation.
This gives you , hence, using the fact that it is possible to take the logarithm of an asymptotic equivalence (this is specific to the logarithm):
.
Using the continuity of the exponential, we deduce:
, hence .
Be careful with this argument: what about hence ? As I wrote, taking the logarithm of an asymptotic equivalence is correct, while taking the exponential is not (except if the argument of the exponential has a limit, like here, in which case the continuity of the exponential function makes it correct).