So, I found on the internet that $(a+b)^n=\sum\limits_{k=0}^n \frac {n!} {k!(n-k)!} a^n b^{n-k}$
Then I started expanding $(a+b+c)^n$ and up to $n=3$ and found a similar pattern, terms that have the highest degrees still have no coefficient (I'm not counting 1) and terms (at $n=3$) with abc in them have coefficients of 3! with terms only containing two variables having coefficients of 2 (2!) and single variable terms having coefficients of 1 (okay, maybe I am counting 1).
This lead me to my question, is there a way to write formula similar to the one for $(a+b)^n$ for things like $(a+b+c)^n$ as well as similar expressions with more variables (like in the title)?
You are looking for the multinomial theorem.
$$(x_1+\cdots+x_m)^n = \sum_{k_1+\cdots+k_m=n}{\frac{n!}{k_1!k_2!\cdots k_m!}x_1^{k_1}x_2^{k_2}\cdots x_m^{k_m}}$$
You can see that when $m=2$ we have that $k_2=n-k_1$ since $k_1+k_2=n$, so this visibly specializes to the binomial theorem in that case.