\begin{align*}
F(x)& = P(X \le x)\\
& = 1 - P(X \gt x)\\
& = 1 - P(\text{first change occurs after an interval of length} x)\\
& = 1 - P(\text{no changes in the interval} [0,x])\\
& = 1 - \frac{(\lambda x)^0 e^{-\lambda x}}{0!}\\
& = 1 - e^{-\lambda x}
\end{align*}
where the discrete Poisson Probability Function is used to answer the probability of exactly no changes in the "fixed" interval [0,x].
Using this distribution function and taking the derivative yields
\begin{equation*}
f(x) = F'(x) = \lambda e^{-\lambda x}.
\end{equation*}