Solve $n < e^{6 \sqrt{n}}$ 
Find for which values of $n \in \mathbb{N}$ it holds that $$n < e^{6 \sqrt{n}}.$$

I tried to use the inequality $(1 + x) \leq e^x$, but from this, I can only find that the inequality holds for $n > 36$. But I need to get $n$ as small as possible.
I also tried the induction on $n$, but I stucked in the induction step. In particular, in showing that $e^{6\sqrt{n}} + 1 \leq e^{6\sqrt{n+1}}$.
I appreciate any help and suggestions.

The inequality holds for all positive $x$, so effectively that a proof seems overkill :-)
(Notice that $0<e^{6\sqrt0}$ and $1<\dfrac6{2\sqrt x}e^{6\sqrt x}$.)

Or using Taylor,
$$n<1+6\sqrt n+18n+\cdots$$
