Q:

What is the sum multiple exponential distributions? 

I have data about the interarrival times of some process.  With the Kolmogorov-Smirnov test I could not reject the hypothesis that the data comes not from an exponential distribution.  So I assumed that the interarrival times are exponential distributed.  Also that my process is a Poisson Point process. 
The process can be divided in multiple classes.  Like a queue, some people have long questions, others are done within seconds.  Some classes have enough data to fit an exponential distribution.  But others have e.g. five data points or less.  I can not fit a distribution with only five observations.  I think it is reasonable to assume all those classes are exponentially distributed.  
Is there mathematical proof that the sum of multiple exponential distributions is exponentially distributed? 

A:

Let $X_1,\ldots,X_n$ be independent with $\mathrm{Expo}(\lambda)$ distribution.  We compute the density of $X_1+X_2$ by convolution:
\begin{align}
f_{X_1+X_2}(t) &= f_{X_1}\star f_{X_2}(t)\\
&= \int_{\mathbb R} f_{X_1}(\tau)f_{X_2}(t-\tau)\ \mathsf d\tau\\
&= \int_0^t \lambda e^{-\lambda \tau}\lambda e^{-\lambda(t-\tau)}\ \mathsf d\tau\\
&= \lambda^2 e^{-\lambda t} \int_0^t\ \mathsf dt\\
&= \lambda ( \lambda t)e^{-\lambda t}\mathsf 1_{(0,\infty)}(t). 
\end{align}
Assuming now that $S_n:=\sum_{k=1}^n X_k$ has density $\frac{\lambda (\lambda t)^{n-1} e^{-\lambda t}}{(n-1)!}$, we compute the density of $S_{n+1} = S_n + X_{n+1}$ again by convolution:
\begin{align}
f_{S_{n+1}}(t) &= f_{S_n}\star f_{X_{n+1}}(t)\\
&= \int_0^t \frac{\lambda (\lambda \tau)^{n-1} e^{-\lambda \tau}}{(n-1)!}\lambda e^{-\lambda (t-\tau)}\ \mathsf d\tau\\
&=\frac{\lambda^2e^{-\lambda t}}{(n-1)!}\int_0^t (\lambda\tau)^{n-1} \ \mathsf d\tau\\
&= \frac{\lambda^2e^{-\lambda t}}{(n-1)!}  \frac{(\lambda  t)^n}{\lambda  n}\\
&= \frac{\lambda(\lambda t)^n e^{-\lambda t}}{n!}, 
\end{align}
so the density is of this form for all positive integers $n$ by mathematical induction.