Let $(X_n)_{n\ge 1}$ iid random variables and $\lambda\ne 0$. Is it true that
$$E[e^{\lambda\sum_{i=1}^nX_i}]=\prod_{i=1}^n E[e^{\lambda X_i}]=\Big(E[e^{\lambda X_1}]\Big)^n,$$
by using independence or maybe another result in the first equation and identically distributed in the second?
Yes, it works exactly like that.
Since the variables $X_i$ for $1 \leq i \leq n$ are independent, and each $e^{\lambda X_i}$ is a function of $X_i$ and no other $X_j$ with $j \neq i$, the variables $e^{\lambda X_i}$ for $1 \leq i \leq n$ are also independent. See, for example, this question and its answers. This gives us
$$ E[e^{\lambda\sum_{i=1}^nX_i}]=E[\prod_{i=1}^n e^{\lambda X_i}]=\prod_{i=1}^n E[e^{\lambda X_i}]. $$
Next, as you already mentioned, we get from identical distribution of the $X_i$ for $1 \leq i \leq n$
$$ E[e^{\lambda X_i}] = E[e^{\lambda X_1}] \qquad for \qquad 1 \leq i \leq n $$
and thus
$$ \prod_{i=1}^n E[e^{\lambda X_i}]=\Big(E[e^{\lambda X_1}]\Big)^n, $$
as desired.
