To begin with, how to show series $a_n=\dfrac{n^{n+\frac{1}{2}}e^{-n}}{n!}, n\in\mathbb{N}^*$ is monotonic and, specifically, increasing?
Now to make it general, replace $\frac{1}{2}$ above with an arbitrary positive parameter $p$. Then when does the series $a_n=\dfrac{n^{n+p}e^{-n}}{n!}$ shows monoticity and if it is monotonic, how to determine if it is increasing or decreasing?
\begin{equation}
a_{n+1}\geq a_n \iff \frac{(n+1)^{n+p+1}}{e^{n+1}(n+1)!}\geq \frac{n^{n+p}}{e^nn!}
\iff \frac{(n+1)^{n+p}}{n^{n+p}}\geq e
\end{equation}
Note that;
\begin{equation}
\left(1+\frac{1}{n}\right)^{n+p} \to e
\end{equation}
Take log;
\begin{equation}
p \geq \frac{1}{\log(1+1/n)} - n \to \frac{1}{2}
\end{equation}
Therefore if $p > 1/2$, then $a_n$'s are increasing, and else they are decreasing, at least eventually.
If you are interested in $p=1/2$, $\log^{-1}(1+1/n)-n$ increases to $1/2$ hence $a_n$'s are increasing.
\begin{equation}\log^{-1}(1+1/n) - n \leq 1/2\end{equation}
Rearrange terms such as;
\begin{equation}
\log(1+1/n) \geq \frac{2}{2n+1} = \frac{1}{n}\left(\frac{1}{1+\frac{1}{2n}}\right)
\end{equation}
Expand both side as $n\to\infty$;
\begin{equation}
\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{kn^k} \geq \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{2^{k-1}n^k}
\end{equation}
Now, you can compare coefficients and note that $k=1,2$ are equal but then $2^{k-1}$ increases exponentially.
When you face problems with factorials, Stirling approximation is very useful. $$a_n=\dfrac{n^{n+\frac{1}{2}}e^{-n}}{n!}\implies \log(a_n)=\left(n+\frac 12\right)\log(n)-n-\log(n!)$$ Using Stirling approximation, we then get $$\log(a_n)=-\frac 12 \log(2\pi)-\frac 1{12n}+O\left(\frac{1}{n^3}\right)$$ which is increasing (have a look at the derivative).
Now, for large values of $n$, use $$a_n=e^{\log(a_n)}=\frac 1 {\sqrt{2\pi}}\left(1-\frac 1{12n}\right)+O\left(\frac{1}{n^2}\right)$$
If you make it more general $$b_n=\dfrac{n^{n+p}e^{-n}}{n!}\implies \log(b_n)=\left(n+p\right)\log(n)-n-\log(n!)$$ and applying the same, we should arrive to $$\log(b_n)=-\frac 12 \log(2\pi)+\left(p-\frac{1}{2}\right) \log (n)-\frac 1{12n}+O\left(\frac{1}{n^3}\right)$$ and then, if $p>\frac 12$, $\log(b_n)$ and then $b_n$ would be increasing.
Since $\log(x)$ is a concave function on $\mathbb{R}^+$, by the Hermite-Hadamard inequality
$$ \sum_{k=1}^{n}\log(n)\leq \tfrac{1}{2}\log(n)+\int_{1}^{n}\log(x)\,dx = 1-n+\left(n+\tfrac{1}{2}\right)\log n$$
and by exponentiating both sides $n!\leq n^{n+\frac{1}{2}}e^{1-n}$ follows.
Let $a_n=\log\left(n^{n+\frac{1}{2}}e^{-n}\right)=(n+\frac{1}{2})\log(n)-n $. We have
$$ a_{n+1}-a_n = \log(n+1)+\left(n+\tfrac{1}{2}\right)\log\left(1+\tfrac{1}{n}\right)-1$$
$$ \frac{d}{dx}\left[\log(x+1)+\left(x+\tfrac{1}{2}\right)\log\left(1+\tfrac{1}{x}\right)-1\right]=\log\left(1+\tfrac{1}{x}\right)-\tfrac{1}{2x(x+1)}\geq 0 $$
hence $\{a_n\}_{n\geq 1}$ is increasing and log-convex. Similarly, by letting
$b_n=\log\left(n^{n+\frac{1}{2}}e^{-n}/n!\right)$ we have
$$ b_{n+1}-b_n = \left(n+\tfrac{1}{2}\right)\log\left(1+\tfrac{1}{n}\right)-1\stackrel{\text{HHI}}{\geq} 0$$
$$ \frac{d}{dx}\left[\left(x+\tfrac{1}{2}\right)\log\left(1+\tfrac{1}{x}\right)-1\right]=\log\left(1+\tfrac{1}{x}\right)-\tfrac{x+\frac{1}{2}}{x(x+1)}\stackrel{\text{HHI}}{\leq} 0 $$
hence $\{b_n\}_{n\geq 1}$ is increasing and log-concave, since
$$ \log\left(1+\tfrac{1}{x}\right)=\int_{x}^{x+1}\frac{dt}{t}\leq \tfrac{1}{2}\left(\tfrac{1}{x}+\tfrac{1}{x+1}\right).$$
