The Marginal Distribution of a Multinomial

The simplest way is "combinatorial/probabilistic." Recall that the "three-nomial" distribution measures the probability of $x_1$ Type 1 events, $x_2$ Type 2 events, and $n-x_1-x_2$ Type 3 events, when an experiment is repeated independently $n$ times, with probabilities of "success" respectively equal to $p_1$, $p_2$, and $p_3$, where $p_1+p_2+p_3=1$.

The probability of $x_1$ Type 1 events is therefore $$\binom{n}{x_1}p_1^{x_1}(p_2+p_3)^{n-x_1}.\tag{1}$$ It follows that the marginal distribution of $X_1$ is binomial.

If we really wish to sum, by the Binomial Theorem the probability (1) is equal to $$\binom{n}{x_1}p_1^{x_1}\sum_{x_2=0}^{n-x_1}\binom{n-x_1}{x_2}p_2^{x_2}p_3^{n-x_1-x_2}.$$ This is precisely the same as the result of summing over all (appropriate) $x_2$ the probability that $X_2=x_2$ and $X_3=n-x_1-x_2$. If we want to save writing down (1) until the very end, we can just write the sum argument backwards.

To see it is relatively easy by the sum of indicators approach.

To verify/proof from the trinomial case:

$$\begin{align} \Pr\{X_1 = x_1\} &= \sum_{x_2=0}^{n-x_1} \Pr\{X_1 = x_1, X_2 = x_2\} \\ &= \sum_{x_2=0}^{n-x_1} \frac {n!} {x_1!x_2!(n-x_1-x_2)!} p_1^{x_1}p_2^{x_2}(1-p_1-p_2)^{n-x_1-x_2}\\ &= \frac {n!} {x_1!}p_1^{x_1} \sum_{x_2=0}^{n-x_1} \frac {1} {x_2!(n-x_1-x_2)!} p_2^{x_2}(1-p_1-p_2)^{n-x_1-x_2} \\ &= \frac {n!} {x_1!(n-x_1)!}p_1^{x_1} \sum_{x_2=0}^{n-x_1} \frac {(n - x_1)!} {x_2!(n-x_1-x_2)!} p_2^{x_2}(1-p_1-p_2)^{n-x_1-x_2} \\ &= \frac {n!} {x_1!(n-x_1)!}p_1^{x_1} [p_2 + (1 - p_1 - p_2)]^{n-x_1} \\ &= \frac {n!} {x_1!(n-x_1)!}p_1^{x_1} (1 - p_1)^{n-x_1}, x_1 = 0, 1, \ldots, n \end{align}$$

So the only key steps are: recognizing the support when $X_1 = x_1$ and recognizing the binomial theorem / expansion.