Show that a random variable follows a gamma distribution

Question

For $n\in \mathbb{N}$, let $T_n$ denote the time when the n-th event occurs. Given that $X_t ∼ Poi(\lambda t)$, where $X_t$ denotes the number of events in the time interval $[0,t]$, I want to show that $T_n ∼\Gamma(n,β)$ where $β = 1/λ$.
My attempt at a solution went as follows
$\mathbb{P}(T_n>t)=\mathbb{P}(X_t=n-1)=\frac{\lambda^{n-1}e^{-\lambda t}}{(n-1)!}$
$\mathbb{P}(T_n\leqslant t)=1-\frac{\lambda^{n-1}e^{-\lambda t}}{(n-1)!}$
$$F_{T_n}(t)=\begin{cases} 1-\frac{\lambda^{n-1}e^{-\lambda t}}{(n-1)!} & \text{if $t>0$} \\ 0 & \text{if $t\leqslant0$} \end{cases}$$
$$f_{T_n}(t)=\begin{cases} \frac{\lambda^{n}e^{-\lambda t}}{(n-1)!} & \text{if $t>0$} \\ 0 & \text{if $t\leqslant0$} \end{cases}$$
Which is close, but not quite what I'm looking for. Could anyone give me a pointer as to where I am going wrong? I know that the end result should look like $$f_{T_n}(t)=\begin{cases} \frac{\lambda^{n}e^{-\lambda t}t^{n-1}}{(n-1)!} & \text{if $t>0$} \\ 0 & \text{if $t\leqslant0$} \end{cases}$$