Using mathematical induction prove the question below : $2^n≤(n+1)!$ For n≥0

Question

I could only go this far but I'm not even sure the method is necessarily right. Step 1.for $n=0$,$1≤1$ which is true

Step 2,assuming its true for $n=k$ $2^k≤(k+1)!$

Step 3,showing it is true for $n=k+1$ $2^{k+1}≤(k+2)!$ as $2^{k+1}≤(k+1)!2\leq(k+1)!(k + 2) = (k + 2)!$

Answer 1

You have assumed that

$$2^k \le (k+1)!$$

We want to show

$$2^{k+1} \le (k+2)!$$

We see that, applying the hypothesis at the point in blue,

$$2^{k+1} = 2 \cdot \color{blue}{2^k} \le 2 \cdot \color{blue}{(k+1)!}$$

We know that $0 \le k$ since we're inducting on $n$ (or $k$ here). Therefore, $2 \le k+2$. Therefore, the previous line implies

$$2 \cdot (k+1)! \le (k+2) \cdot (k+1)!$$

We recall that $m! = m\cdot (m-1)!$ by the definition (provided $m$ is a positive integer). Therefore,

$$(k+2) \cdot (k+1)! = (k+2)!$$

Therefore, in summary,

$$\begin{align} 2^{k+1} &= 2 \cdot 2^k \\ &\le 2 \cdot (k+1)! \\ &\le (k+2) \cdot (k+1)! \\ &= (k+2)! \end{align}$$

completing the induction step, as it verifies that which we sought to prove.

Answer 2

We can prove this by induction.

Assume $k\geq1$ and $2^{k-1}\leq k!$. Then, we have $2\leq k+1$. We can multiply the first and second inequalities to get $2\cdot2^{k-1}\leq (k+1)\cdot k!$ which is equivalent to the original statement $$2^k\leq (k+1)!$$

All we need to do now is show $2^k\leq (k+1)!$ holds for $k=0$ (which you have already done).