Proof
Take the derivative of the probability function to get
\begin{equation*}
f'(x) = \frac{\sqrt{2} {\left(\mu - x\right)} e^{\left(-\frac{{\left(\mu - x\right)}^{2}}{2 \, \sigma^{2}}\right)}}{2 \, \sqrt{\pi} \sigma^{3}}
\end{equation*}
which is zero only when \(x = \mu\text{.}\)
It is easy to see by evaluating to the left and right of this value that this critical value yields a maximum.