Help solving the inequality $2^n \leq (n+1)!$, n is integer

Question

I need help solving the following inequality I encountered in the middle of a proof in my calculus I textbook:

$2^n \leq (n+1)!$

Where $\mathbf{n}$ in an integer.

I've tried applying lg to both members, but got stuck at:

$n \leq \lg(n+1) + \lg(n) + \lg(n-1) + ... + \lg(3) + 1$

A proof by induction is acceptable, but I wanted an algebraic one. I find it more... elegant?

Answer 1

$(n+1)!=2\cdot 3\cdot \dots\cdot(n+1)$ here a product of $n$ numbers all are at least 2 so the result follows...

Answer 2

HINT:

$$\frac{(n+1)!}{2^n}=\frac{2}{2}\frac{3}{2}\frac{4}{2}\cdots \frac{n-1}{2}\frac{n}{2}\frac{n+1}{2}$$