Proof
For a sufficiently large natural number n, break up the given interval [0,T] into n uniform parts each of width h = T/n. Using the properties of Poisson processes, n very large implies h will be very small and eventually small enough so that
\begin{equation*}
P(\text{exactly one success on a given interval}) = p = \lambda \frac{T}{n}.
\end{equation*}
However, since there are a finite number of independent intervals each with probability p of containing a success then you can use a Binomial distribution to evaluate the corresponding probabilities so long as n is finite. Doing so yields and taking the limit as n approaches infinity gives:
\begin{align*}
f(x) & = P(x \text{changes in} [0,T]) \\
& = \lim_{n \rightarrow \infty} \binom{n}{x} p^x (1-p)^{n-x}\\
& = \lim_{n \rightarrow \infty} \binom{n}{x} (\frac{\lambda T}{n})^x (1-\frac{\lambda T}{n})^{n-x}\\
& = \lim_{n \rightarrow \infty} \frac{n(n-1)...(n-x+1)}{x!} ( \frac{\lambda T}{n})^x (1- \frac{\lambda T}{n})^{n-x}\\
& = \frac{(\lambda T)^x}{x!} \lim_{n \rightarrow \infty} \frac{n(n-1)...(n-x+1)}{n \cdot n \cdot ... \cdot n} (1-\lambda \frac{T}{n})^{n}(1-\lambda \frac{T}{n})^{-x}\\
& = \frac{(\lambda T)^x}{x!} \lim_{n \rightarrow \infty} (1-\frac{1}{n})...(1-\frac{x-1}{n}) (1- \frac{\lambda T}{n})^{n}(1- \frac{\lambda T}{n})^{-x}\\
& = \frac{(\lambda T)^x}{x!}
\lim_{n \rightarrow \infty} (1- \frac{\lambda T}{n})^{n}
\lim_{n \rightarrow \infty} (1- \frac{\lambda T}{n})^{-x}\\
& = \frac{(\lambda T)^x}{x!}
\lim_{n \rightarrow \infty} (1- \frac{\lambda T}{n})^{n} \cdot 1\\
& = \frac{(\lambda T)^x}{x!}
e^{-\lambda T}
\end{align*}
where L’Hopitals rule was utilized in the final step.