In converting between two definitions of a Poisson process, namely starting from the "exponential inter arrival-times" definition and attempting to prove the "Poisson distribution of point counts" definition, I believe my lecture notes might be overcomplicating things.
We are given the definition of a Poisson process as $X_t= |\{{n\geq 1:\sum_{k=1}^nZ_k\leq t}\}|$, with $Z_i\sim Exp(\lambda)$ i.i.d. and the aim is to prove that $X_t \sim Poi(\lambda t)$, i.e. $\mathbb{P}(X_t = k) = \frac{(\lambda t)^k e^{- \lambda t}}{k!}$.
My notes use the fact that the sum of these exponential random variables has a $Gamma(n,\lambda)$ distribution, and then integrates by parts on the gamma density. Fearing for my ability to accurately memorise the gamma density function, what is wrong with the following approach?
$$\mathbb{P}(X_t = k) = \frac{(\lambda t)^k e^{- \lambda t}}{k!} = \frac{\lambda t}{k} \bullet \frac{(\lambda t)^{k-1} e^{- \lambda t}}{(k-1)!} = \frac{\lambda t}{k} \bullet \mathbb{P}(X_t = k-1)$$
Therefore, given the base condition $\mathbb{P}(X_t = 0) = \mathbb{P}(Z_1 > t) = e^{-\lambda t}$, if we can prove the relation $\mathbb{P}(X_t = k) = \frac{\lambda t}{k} \bullet \mathbb{P}(X_t = k-1)$ then we get an easy gamma-free proof.
Can anyone see why this relation might hold? Or any comments on the validity of this proof.