Induction proof, greater than Prove that: $n!>2^n$ for $n \ge 4$.
So in my class we are learning about induction, and the difference between "weak" induction and "strong" induction (however I don't really understand how strong induction is different/how it works.
Here's my attempt:
Let S(n) be the statement $n!>2^n$ for $n \ge 4$.
Then let n=4
$$4!>2^4$$
$$24>16$$
so S(1) is true. Next, suppose that k is any integer with $k \ge 4$ such that S(k) is the statement $k!>2^k$. Then, multiplying (k+1) to both sides:
$$k!>2^k$$
$$k!(k+1)>2^k(k+1)$$
$$(k+1)!>k2^k+2^k$$
And this is where I am stuck....
S(k+1) is the statement $(k+1)!>2^(k+1)=2*2^k$ so then $k!(k+1)>2^k(k+1)>2*2^k$ because $k \ge 4$? Is this the right thinking, and if so, I am still not really sure how to formally write this out.
 A: You're so close... all you need is $$ (k+1)!>k2^k+2^k>2^k+2^k=2^{k+1}$$
Earlier you lay out your demonstraion of the inital induction condition slightly awkwardly; I would say: $$ 4!=24>16=2^4$$ which reading left to right makes unremarkable  claims at each step and finishes with the required inequality.
A: For $n\geq 4$, denote the statement involving $n$ by
$$
S(n) : 2^n<n!.
$$
Base step ($n=4$): Since $2^4=16$ and $4!=24$, the statement $S(4)$ is true.
Inductive step: Fix some $k \geq 4$ and assume that
$$
S(k) : 2^k < k!
$$
is true. To be shown is that 
$$
S(k+1) : 2^{k+1} < (k+1)!
$$
follows. Beginning with the left side of $S(k+1)$,
\begin{align}
2^{k+1} &= 2(2^k)\\[0.5em]
        &< 2(k!)\tag{by $S(k)$}\\[0.5em]
        &< (k+1)(k!)\tag{since $k\geq 4$}\\[0.5em]
        &= (k+1)!,
\end{align}
the right side of $S(k+1)$. This concludes the inductive step $S(k)\to S(k+1)$.
Thus, by mathematical induction, for all $n\geq 4$, the inequality $S(n)$ is true.
A: Hint: Notice that $k 2^k > 2^k$, for $k \geq 2$.
