Proving $\binom{2n}{n}\le 4^n$ for all $n$ by smallest counterexample 
Prove $$\binom{2n}{n}\le 4^n$$ for all natural numbers $n$ by smallest (minimal) counterexample.

My attempt:
First, $$\binom{2n}n = \frac{(2n)!}{(n!)^2} \le 4^n\;.$$ We know that $x\ne 0$ because $\frac{(2\cdot 0)!}{(0!)^2} = 1$ which is true. So $x\ge 2$. Now consider $x-1\in \Bbb N$. Also note that $x-1 <x$ and is the smallest counterexample. So, $n=x-1$.
$$\frac{(2(x-1))!}{((x-1)!)^2} \le 4^{x-1}$$
$$\frac{(2x-2)!}{((x-1)!)^2} \le 4^{x-1}$$
So this is where I'm stuck. Do I keep on expanding? If so, how?

HINT: In order to complete your proof, you need to show that the inequality $$\frac{(2x-2)!}{((x-1)!)^2} \le 4^{x-1}\tag{1}$$ implies that $$\frac{(2x)!}{x!^2}\le 4^x\;.$$
Now $$\frac{(2x)!}{x!^2}=\frac{2x(2x-1)(2x-2)!}{x^2(x-1)!^2}=\frac{2x(2x-1)}{x^2}\cdot\frac{(2x-2)!}{(x-1)!^2}\le\frac{2x(2x-1)}{x^2}\cdot4^{x-1}$$ by virtue of the hypothesis $(1)$. You’ll be done if you can show that
$$\frac{2x(2x-1)}{x^2}\le 4\;.$$
Can you finish it now.
