Prove $\sqrt{n!} \gt \frac{n}{2}$ $\forall$ $n \ge 1$ Prove that $$\sqrt{n!} \gt \frac{n}{2}$$ $\forall$ $n \ge 1$
My try:
I Tried using Induction:
$P(1)$ is True obviously.
Let $P(k)$ Be True Then we have
$$\sqrt{k!} \gt \frac{k}{2}$$
Now $$\sqrt{(k+1)!}=\sqrt{k+1}\sqrt{k!}\gt \sqrt{k+1}\frac{k}{2}$$
Now Since $x^2-x-1 \gt 0$ $\forall $ $x \ge 2$ We have
$$k^2 \gt k+1$$ $\forall$ $k \ge 2$
Hence
$$k \gt \sqrt{k+1}$$
Hence
$$\sqrt{(k+1)!} \gt \sqrt{k+1}\frac{k}{2}\gt \frac{1}{2}\sqrt{k+1}\sqrt{k+1}=\frac{k+1}{2}$$
Hence Proved
Is there any alternate approach?

Assume $n \geq 2$,
Then, $$\frac{4}{3} < n$$
$$\Leftrightarrow n < 4(n-1)$$
$$\Leftrightarrow \frac{n}{4} < (n-1).$$
Then 
$$\frac{n}{4} < (n-1) \leq (n-1)!$$
Multiply by $n$ and take the square root
$$ \frac{n}{2} < \sqrt{n!}$$
