The binomial expansion tells us how to write the expression (a+b)^{N}
in terms of a sum of various powers of a and b. The way to do this in principle
is just to write out all the factors and expand term by term. For example

(a+b)^{2} = (a+b) x
(a+b) = a x (a+b) + b x (a+b)

= aa + ab + ba + bb

= a^{2} + ab + ab +b^{2}
(note: this expression has 4 terms)

= a^{2} + 2ab + b^{2}

where ab = a x b, and from now on we will stop using the "x" symbol for multiplication, and just indicate multiplication by writing the two quantities next to each other.

(a+b)^{3} = (a+b)(a+b)(a+b)
= a(a+b)(a+b) + b(a+b)(a+b)

= aa(a+b) + ab(a+b) + ba(a+b) + bb(a+b)

= aaa + aab + aba + abb + baa + bab + bba + bbb

= a^{3} + a^{2}b
+ a^{2}b + ab^{2}
+ a^{2}b + ab^{2}
+ ab^{2} + b^{3
}(note: the above expression has 8 terms)

= a^{3} + 3a^{2}b
+ 3ab^{2} + b^{3}

In general,

(a+b)^{N} = (a+b)(a+b)(a+b)...(a+b)

where "..." means there are a total of N such terms.
When we expand out, each resulting term can be viewed as if it were the
result of "choosing" either an "a" or a "b" from each of the N factors
(a+b). Since there are two such choices for each factor (a+b), we see that
we will wind up with 2^{N} terms (check that
this agrees with the above cases: for N=2 we found 2^{2}=4
terms, for N=3 we found 2^{3}=8 terms).

Consider now the particular case where we have chosen
"b" r times and hence "a" is chosen the remaining N-r times, we wind up
with the term a^{N-r}b^{r}.
There will be many such terms a^{N-r}b^{r}
for a given value of r, as there are many different ways to choose the
r "b"s from the N factors (a+b). How many ways are there? There are exactly
^{N}C_{r}
ways! So when we expand (a+b)^{N}, the term
a^{N-r}b^{r}
appears
times. We thus can group them all together into a single term:

Now there will such terms for all values of r = 0,
1, 2, ..., N. Note that for the term r=0 we choose "b" no times, and choose
"a" N times. The resulting factor, a^{N},
is consistent with the above formula provided we regard 0!=1, as we said
earlier in our definition of the factorial, and also b^{0}=1.
This last result b^{0}=1 is consistent with
the usual mathematical conventions of power laws. Similarly, for the term
r=N, the above formula is consistent with b^{N}.

So finally we have the __binomial expansion__:

Where we use the fact that x^{0}
= 1, and x^{1} = x, for any number x. You
can check that this formula gives the correct results for the cases N=2
and N=3 which we did explicitly above.