Let $f(n)=n!$ and $g(n)=\binom{2n}{n}$. Obviously $f(7)\gt g(7)$ and $f(6)\lt g(6)$. The functions are increasing on $n\in (1,\infty)$, $n$ being an integer. But this is not enough to prove $$n!\gt\binom{2n}{n}\iff n\ge 7.$$
Maybe convexity could help. We could use the fact that $n!$ is convex but I don't know how to prove the convexity of $\binom{2n}{n}=\frac{(2n)!}{n!^2}$ from this.
You can use $$ \frac{{(2n)!}}{{n!^2 }} < \frac{{4^n }}{{\sqrt {\pi n} }},\quad n! > \left( {\frac{n}{e}} \right)^n \sqrt {2\pi n} $$ for all $n\geq 1$. Then it is enough to show $$ 1 < \left( {\frac{n}{{4e}}} \right)^n \pi \sqrt 2 n $$ for $n\geq 7$, which is true. (Trivially true for $4e <11 \leq n$ and the cases $n=7,8,9,10$ can be checked by hand.)
You can also argue that $$ \binom{2n}{n} \le 4^n < n!. $$ for $n\geq 9$.