Expanding brackets of the form $(a+b)^n$ If we have the equation
$
(a+b)^n = (a+b)\times(a+b)\times(a+b)\times \ldots \times (a+b)
$
expanding the right hand side of the above, is the sum of terms in the form $a^n$, $a^{n-1}b$,$a^{n-2}b^2$, $\ldots$, $b^n$. That is $a^{n-1}b^k$ for $k=0,1,\ldots,n$
Now, my textbook states the following

Each term of the form $a^{n-k}b^k$ arises by choosing the variable $b$
  from $k$ of the pairs of brackets on the right hand side of the
  equation and the variable $a$ from the remaining $n-k$ pairs of
  brackets.

My question is now, why? There is no proof given for this, and I am struggling to understand why the above holds.For a very simple expansion, where $k=1$ and $n=2$, we have 2 terms of the form $ab$, here its pretty easy to see from
$
(a+b)(a+b) = a(a+b) + b(a+b) = a^2 + b^2 + 2ab
$
why the statement in the book holds for this case. But, for bigger ones, such as $(a+b)^3$ i can't see it. I can see that there is 3 ways of choosing one $b$ and two $a$, for the term $a^2b$ when looking at $(a+b)(a+b)(a+b)$, but I cant relate it to the expansion of them, if that makes sense?
 A: To simplify talking about the statement a bit, let $C_i = (a + b)$ for $1\le i \le b$, so
$$(a + b)^n = C_1 \times \dots \times C_n.$$
For the case where $n = 3$ and the term $a^2b$:
If we choose $b$ from $C_1$, then we choose $a$ from $C_2$ and $C_3$.
If we choose $b$ from $C_2$, then we choose $a$ from $C_1$ and $C_3$.
If we choose $b$ from $C_3$, then we choose $a$ from $C_1$ and $C_2$.
Thus the coefficient of $a^2b$ is $3 = \binom{3}{1}$.
More generally you should be able to see that a term of $a^{n - k}b^k$ must have taken either $a$ or a $b$ from any term $C_i$. Thus simply follows from
$$(a + b)^n = C_1 \times\dots\times C_n = (a + b)[C_1 \times\dots C_{i - 1} \times C_{i + 1} \times C_n]$$
Since each $C_i$ contributes exactly one $a$ or $b$, we end up with exactly $k$ of the choices being $b$ and exactly $n - k$ of the choices being $a$. Since there are $\binom{n}{k}$ ways to choose to either take $b$ or not to take $b$ from each $C_i$, we end up with $\binom{n}{k}$ terms of $a^{n-k}b^k$.
