Require assistance proving $n≥2 \Longrightarrow \frac{n!}{n^n} ≤ \frac{1}{2}^{\lfloor \frac{n}{2}\rfloor}$ Theorem: $n≥2 \Longrightarrow \frac{n!}{n^n} ≤ \frac{1}{2}^{\lfloor \frac{n}{2}\rfloor}$
Attempted Solution: 
We use induction. Additionally, we prove the stronger inequality omitting the floor function. That is, $$n≥2 \Longrightarrow \frac{n!}{n^n} ≤ \frac{1}{2^\frac{n}{2}}$$
The base case is clear. Suppose it holds for some $n$. Then, consider the case for $n+1$. On the right side, we get $\frac{1}{2^\frac{n+1}{2}} = \frac{1}{2^{\frac{n}{2} + \frac{1}{2}}} = \frac{1}{2^\frac{n}{2}} \cdot \frac{1}{\sqrt{2}}$. This implies the right side is multiplied by $\frac{1}{\sqrt{2}}$ for each successive increase in $n$; hence we must show the left side is multiplied at most by $\frac{1}{\sqrt{2}}$.
Consider now the left side for $n+1$. It is $\frac{(n+1)!}{(n+1)^{(n+1)}} = \frac{n!}{(n+1)^{n}}$. This implies that for each successive increase in $n$, the left side is being multiplied by $\frac{n^n}{(n+1)^n}$. 
So we must show $\frac{n^n}{(n+1)^n} < \frac{1}{\sqrt{2}}$ for all $n≥2$. It is clear for $n=2$ and it suffices to show the left hand side is strictly decreasing as $n$ increases. I am unable to do this, though Wolfram seems to agree with me. Note that I don't want to differentiate, since the author of the book I am reading has not yet introduced derivatives. 

Another approach is to show that
$$\left(1+\frac1n\right)^{2n}=\left(\frac{n+1}{n}\right)^{2n}>2.$$
The inequality follows directly from the binomial theorem.
