Distribution of the sample mean of a exponential

Correction.

In discussing this question, I have discovered errors here.

Specifically if $n$ observations are sampled at random from $\mathsf{Exp}(\text{rate} = \lambda),$ as shown in the Question above, then $T \sim \mathsf{Gamma}(\text{shape}=n,\, \text{rate}=\lambda).$

The proof is that the MGF of $X_i$ is $M_X(t) = \frac{\lambda}{1-t},$ so the MGF of $T$ is $M_T(t) = (\frac{\lambda}{1-t})^n,$ which is the MGF of $\mathsf{Gamma}(\text{shape}=n,\, \text{rate}=\lambda).$

Consequently, $\bar X \sim \mathsf{Gamma}(n, n\lambda).$ (This relationship is illustrated in the link.)