[proofplan]
We write $M_{X+Y}(\theta) = \mathbb{E}[e^{\theta(X+Y)}]$, factor the exponential as $e^{\theta X} \cdot e^{\theta Y}$, and use the independence of $X$ and $Y$ to split the expectation into a product. This gives $M_X(\theta) \cdot M_Y(\theta)$.
[/proofplan]
[step:Factor the exponential and apply independence]
By definition of the moment generating function,
\begin{align*}
M_{X+Y}(\theta) = \mathbb{E}[e^{\theta(X+Y)}] = \mathbb{E}[e^{\theta X} \cdot e^{\theta Y}].
\end{align*}
Since $X$ and $Y$ are independent, the random variables $e^{\theta X}$ and $e^{\theta Y}$ (being measurable functions of independent random variables) are independent. Both are non-negative, so the expectation of their product equals the product of their expectations:
\begin{align*}
\mathbb{E}[e^{\theta X} \cdot e^{\theta Y}] = \mathbb{E}[e^{\theta X}] \cdot \mathbb{E}[e^{\theta Y}] = M_X(\theta) \cdot M_Y(\theta).
\end{align*}
[/step]