Determine the distribution of the sum of n independent identically distributed Poisson random variable $X_i$ ~ $\mathsf{Poisson}(\lambda)$, $ i = 1, ...., n$?
My approach is that since it is independent identically and the sum is basically $X_1 + X_2 + ... + X_n$ we can just use moment generate function to find it.
So $M_x (t) = E e^{xt} = e^{-\lambda} * e^{e^{\lambda*t} },$ which is just basically that to the $n$ power.
Can anyone confirm that I'm on the right path?

Your technique will work, and is a good exercise to help learn to work with generating functions.
The easy way to do the problem, however, is to note that a Poisson distribution represents the number of events in a fixed interval of time, given that the events occur with a given fixed average rate. The rate, in events per second, of the sum of these will be the sum of the rates. Thus if all the independent $X_i$ have $\lambda = \lambda_1$ the answer will be a Poisson distribution with $$ \lambda = \sum \lambda_i = n\lambda_1 $$