TITLE: Showing n! is greater than n to the tenth power
QUESTION [21 upvotes]: I'd like to show $n!>n^{10} $ for large enough n ( namely $ n \geq 15 $). 
By induction, I do not know how to proceed at this step: 
$$ (n+1)\cdot n!>(n+1)^{10}  $$
As I can't see how to simplify $(n+1)^{10} $. 
This seems like such a trivial thing (and it probably is), yet I can't do it. Isn't there an easier way to show this? 
(P.S. I need to refrain from the use of derivatives, integrals etc., I suppose, then you could work something out with the slope of the respective functions)

REPLY [3 votes]: There's no need for induction or anything beyond elementary arithmetic. For $n\geq 15$,
\begin{align*}
    n!&\geq n(n-1)\cdots(n-9)\times 5\cdot 4\cdot 3\cdot 2 \\
      &= 120n(n-1)\cdots(n-9)\\
      &> n \cdot \frac{15}{14}(n-1)\cdot \frac{15}{13}(n-2)\cdots \frac{15}{6}(n-9)\\
      &\geq n^{10}\,,
\end{align*}
where the strict inequality is because the third line is approximately $53n(n-1)\cdots(n-9)$ and the final inequality is because, for all $i$ and $n$ with $1\leq i<15\leq n$,
$$\frac{15}{15-i}(n-i) = n + \frac{(n-15)i}{15-i} \geq n\,.$$