The sum $$(a+b)(a+b)...(a+b) $$
$$=\sum_{t_j \in \{a,b\}} t_1t_2,...,t_n $$
= $$ \sum_{j}{ n \choose {j,n-j}} a^jb^{n-j}$$ = $$ \underbrace{aa...a}_{n} + \underbrace{aa...a^b}_{n-1}+...$$
and $\sum_{j=0}^n {n \choose j} = 2^n$
How are there $2^n$ terms in this sum?
Similarily
$$\sum_{t_j\in \{a,b,c\}} t_1t_2t_3t_4t_5=aaaaa+...+ccccc$$
There are $3^5$ terms in the sum: Why?
Let's have $n = 3$ for illustrative purposes. Then we should expect $2^3 = 8$ terms. And indeed, we get $$ (a+b)^3 = aaa + aab + aba + baa + abb + bab + bba + bbb $$ for a total of $8$ terms.
The binomial theorem then exploits the fact that multiplication is assumed to be commutative, which gives us $aab = aba = baa$ and $abb = bab = bba$, so we can rewrite the above sum into $$ aaa + aab + aab + aab + abb + abb + abb + bbb\\ = aaa + 3aab + 3abb + bbb $$ This time with is just four terms, as terms that have equal value (that have the same number of $a$'s and the same number of $b$'s, regardless of order) are grouped toghether and multiplied by the number of terms in each grouping (the binomial coefficients). Include exponents on top of this, and you have the conventional expansion: $$ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 $$ This all generalizes neatly to all natural numbers $n$.