When is $n!>x^n$? When is $n!>x^n$? assuming that x is a fixed positive. I know we can take the log of both sides and use the following formula:
$$n\log(x) = \log(x^n) < \log(n!) = \sum_{i = 1}^n\log(i)$$.
But this still gives me difficulty trying to find a specific N that makes the right side larger. Is there another formula I should be using?
 A: Use that (see here) for $n\ge 1$:
$$
n!>\sqrt{2\pi n}\left(\frac{n}{e}\right)^ne^{\frac{1}{12n
+1}}.
$$
Hence it is enough to check that
$$
\sqrt{2\pi n}\left(\frac{n}{e}\right)^ne^{\frac{1}{12n
+1}}\ge x^n,
$$
where $x$ is fixed and positive. Equivalently
$$
2\pi n\left(\frac{n}{ex}\right)^{2n}e^{\frac{2}{12n
+1}}\ge 1.
$$
This holds, of course, for all $n\ge n_0(x):=ex$.
Edit: As pointed out in the comments below by SomeCallMeTime, this estimate is asymptotically optimal.
A: There is an almost exact solution for the equation $$n!=x^n\tag 1$$ Have a look at @robjohn's answer to this question of mine.
Adapted to $(1)$ (that is to say $a=x$ and $k=0$ in my post), it gives (as a real)
$$n\sim x\, e^{1+W(t)}-\frac 12 \quad \text{where}\qquad t=-\frac{\log (2 \pi  x)}{2 e x}\tag 2$$ $W(t)$ being Lambert function.
Suppose $x=1234$. The above will give $n=3349.378835$ while the exact solution is $3349.378848$. As usual, you will use $\lceil n \rceil$.
If $x$ is "large", $t$ is small and you could use the approximation
$$W(t) \sim t \, \frac{1+\frac{19 }{10}t+\frac{17 }{60} t^2} {1+\frac{29 }{10}t+\frac{101 }{60}t^2 }+O\left(t^6\right)$$
Using it for the worked case, it would give ... the same.
Edit
If you want a shortcut approximation for $n \leq 1000$, you could use the empiriccal
$$n  \approx a\,x^b-c$$ obtained by a quick and dirty regression $(R^2>0.999999)$.
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & 2.68239 & 0.00045 & \{2.68151,2.68327\} \\
 b & 1.00195 & 0.00003 & \{1.00190,1.00201\} \\
 c & 2.53581 & 0.01065 & \{2.51491,2.55671\} \\
\end{array}$$
It gives a maximum error of $\sim 0.1$ in the range.
A: If you need only estimate for $n$, you can use following claim, which is simple to prove: $n! \geq (\sqrt{n})^n$.
$$(n!)^2=(1\cdot 2\cdot ...\cdot n)\cdot(n\cdot(n-1)\cdot...\cdot 1)
=(1\cdot n)\cdot(2\cdot(n-1))\cdot...\cdot(n\cdot 1)$$
Every product is of form
$$k\cdot(n+1-k)=nk-k(k-1)=n+n(k-1)-k(k-1)=n+(n-k)(k-1)\geq n$$
There are exactly $n$ such products, then
$$(n!)^2 \geq n^n \Rightarrow n! \geq (\sqrt{n})^n$$
So for $n > x^2$ one can be sure that $n! \geq (\sqrt{n})^n > x^n$.
