Induction proof $n!>2n$ for all integers $n>3$

Question

I'm trying to solve it, but I'm stuck with

$$(k+1)k!>2(k+1)$$

Answer 1

Base case: $4!=24 > 8 = 2\cdot 4$

Induction assumption: $k>3$ and $k!>2k$

$$\begin{align} (k+1)! &= (k+1)k! \\ &>(k+1)2k \tag{ind. hyp}\\ &>2(k+1) \\ \end{align}$$

Therefore the induction holds and $n!>2n$ for $n>3$.

As you can see this is a very loose lower bound on $n!$. We could achieve $n!>5n$ in essentially the same argument.

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In fact that is so loose I'm wondering if the original intent was to claim that $n!>2^n$ for $n>3$. In which case the induction goes:

Base case: $4!=24 > 16 = 2^4$

Induction assumption: $k>3$ and $k!>2^k$

$$\begin{align} (k+1)! &= (k+1)k! \\ &>(k+1)2^k \tag{ind. hyp}\\ &>2^{k+1} \tag{$k\mathord+1>2$} \\ \end{align}$$

Answer 2

According to our induction, we should have

$$(k+1)k!>(k+1)(2k)>2(k+1)$$

since we assume $k!>2k$. Lastly, we check $k=4$.