Let $k \geq 6$ and I know $k!$ < $\dfrac{k^k}{2^k}$ I want to show the following:
$(k+1)! < \dfrac{(k+1)^{k+1}}{2^{k+1}}$
Now I am going to show my solution, let me know if my reasoning is correct as well.
Simplifying what we have above we get: $(k+1)k!$ < $\dfrac{(k+1)(k+1)^k}{2^{k+1}}$
Now divide $(k+1)$ from both sides: $k!$ < $\dfrac{(k+1)^k}{2^{k+1}}$
Next, factor out a k from the right side: $k!$ < $\dfrac{k^k(1+ \frac{1}{k})^k}{2^{k+1}}$
Now we know $\lim_{k\to\infty}$ $(1+ \frac{1}{k})^k$ is $e$ so we can write:
$k!$ < $\dfrac{k^ke}{2^{k+1}}$ $\iff$ $k!$ < $\dfrac{k^ke}{2\cdot2^k}$
This is also equivalent to: $k!$ < $\dfrac{k^k}{2^k}$ $\cdot$ $\dfrac{e}{2}$
Since I know $k!$ < $\dfrac{k^k}{2^k}$ (from an induction hypothesis) and I am multiplying by $\dfrac{e}{2}$ which is > $1$ the inequality holds. End of Proof.
The step I am most unsure of is setting the $(1+ \frac{1}{k})^k$ to $e$. Can I do this?