Prove that the function $f(n) = n! - 2^n$ is positive for $n \ge 4$

Question

1. n ∈ N and $P(n) : n! − 2^n > 0$.
2. $P(4) : 4! − 16 > 0$ is true.
3. $P(m)$ is true, m ≥ 4.
4. $m! − 2^m > 0$, from step 3.
5. $(m+1)! − 2^{m+1} = (m+1)\cdot m! − 2\cdot2^m$.
6. $m+1 > 2$, from step 3.
7. $(m+1)(m! − 2^m)>m! − 2^m$, from steps 4 and 5.
8. $P(m+1)$ is true, from steps 4 and 6.
9. $P(m)⇒P(m+1)$.
10. From steps 2 and 9 infer P(n) is true.

I don't understand step 5. $(m+1)\cdot m! − 2\cdot 2^m$.

Answer 1

$$\begin{align} (m+1)!&=(m+1)\cdot m\cdot(m-1)\cdots2\cdot1\\ &=(m+1)\cdot m!\\ \end{align}$$

And $$\begin{align} 2^{m+1} &=\underbrace{2\cdot2\cdots2\cdot2}_{m+1 \text{times}}\\ &=2\cdot\underbrace{2\cdot2\cdots2\cdot2}_{m \text{times}}\\ &=2\cdot2^m\\ \end{align}$$

Answer 2

Hint:-

$\left((m+1)!-2^{m+1}\right)-\left(m!-2^{m}\right)=m!(m+1)-2^m>0$

Answer 3

Step 5 is trying to write the result for m+1 in terms of the result for m so that induction can be used.