Proof
For the mean, use integration by parts with \(u = x\) and \(dv = \lambda e^{-\lambda \cdot x}\) to get (eventually)
\begin{align*}
\text{Mean} = \mu & = \int_0^{\infty} x \cdot \lambda e^{-\lambda x} dx\\
& = \left [ -(x+\frac{1}{\lambda}) e^{-\lambda \cdot x} \right ] \big |_0^{\infty} \\
& = \frac{1}{\lambda}.
\end{align*}
and so the use of \(\lambda = \frac{1}{\mu}\) in f(x) is warranted.
The remaining statistics are derived similarly using repeated integration by parts. The interactive Sage cell below calculates those for you automatically.