Induction proof: $n! > 2^n$ for all $n \geq 4$

Question

Prove: $n! > 2^n$ for all $n \geq 4$

I already proved the base case: $24 > 16$.

Then I assume this holds for $n=k$, and start proving it for $n=k+1$:

$(k+1)k! > 2^{k+1}$

$(k+1)k! > 2^{k}(2^1)$

After this I'm not sure how to proceed. Any help is appreciated, thanks in advance.

Answer 1

Well, notice that

$$ (k+1)! = (k+1) k! \geq (k+1) 2^k = k 2^k + 2^k > 2^k+2^k=2\cdot2^k= 2^{k+1} $$

Answer 2

The induction hypothesis is that $k!>2^k$ for some $k\geq4$. Then $(k+1)! =k!(k+1) > 2^k(k+1)$. But $2^k(k+1)>2^k2$, since $k\geq 4$, and so $k(k+1)!>2^{k+1}$.

Answer 3

Step :

$2(n!) >2 \cdot 2^n=2^{n+1}$(hypothesis).

For $n \ge 4:$

$(n+1)!= (n+1)n! > 2n! > 2^{n+1}.$