Use stirlings approximation to prove inequality.

I have come across this statement in a text on finite elements. I can give you the reference if that will be useful. The text mentions that the inequality follows from Stirling's formula. I can't prove it to myself but I think it is true from checking values with Mathematica.

Let $p,k\in \mathbb{N}$ with $k\le p$. Then we have

$$\frac{(p-k)!}{(p+k)!} \le \left(\frac{\theta}{p}\right)^{2k},\, \theta = \left(\frac{e}{2}\right)^{k/p}.$$

Does anyone have an idea of how to get this inequality from Stirling's approximation?

This is my attempt at a solution. To build intuition, let's look at the case $k=p$ we have

$$\begin{align*} \frac{(p-k)!}{(p+k)!} &= \frac{1}{(2p)!} \\ &\le \frac{1}{\sqrt{2\pi}} \frac{\exp(2p)}{(2p)^{2p+1/2} } \\ &= \frac{1}{\sqrt{2\pi}} \left(\frac{e}{2} \right)^{2p} \frac{1}{\sqrt{2}} \frac{1}{p^{2p}} \frac{1}{p^{1/2}}\\ & \le \frac{1}{\sqrt{2\pi}} \left( \frac{e}{2} \right)^{2p} \frac{1}{\sqrt{2}} \frac{1}{p^{2p}} \\ &= \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{2}} \left( \frac{\theta}{p} \right)^{2k}\\ &\le \left( \frac{\theta}{p} \right)^{2k} \end{align*}$$

And this is what we wanted to show.

Now let's look at the case $k=p-1$, then we get

$$\begin{align*} \frac{(p-k)!}{(p+k)!} &= \frac{1}{(2p-1)!} \\ &\le \frac{1}{\sqrt{2\pi}} \frac{\exp( 2p-1)}{ (2p-1)^{2p-1/2}} \\ &= \frac{1}{\sqrt{2\pi}} \frac{\exp(2p-1)}{2^{2p-1/2}} \frac{1} {(p-1/2)^{2p-1/2} }\\ & = \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{2}} \left(\frac{e}{2} \right)^{2p-1} \frac{1}{(p-1/2)^{2p-1/2}}\\ &\le \frac{1}{2\sqrt{\pi}} \left( \frac{e}{2} \right)^{2p-1} \frac{1}{(p-1/2)^{2p-1}} \\ &\le ?? \end{align*}$$

This is where I am stuck. For the $k=p-1$ case, I can multiply and divide by $(\frac{e}{2})^{3-2/p}$ to get something of the correct form, but then it seems I would have to to prove that for all $p$

$$\left(\frac{e}{2}\right)^{3-2/p} \frac{1}{(p-1/2)^{2p-1}} \le 1.$$ This seems like a dead end to me. And this is still just a special case of the general result. Any ideas?