Question 1002501
{{{(x+y)^n=x^n+nx^(n-1)y+(n(n-1)/2)x^(n-2)y^2+"..."+(n(n-1)/2)x^2y^(n-2)+nxy^(n-1)+y^n}}}
is a formula/theorem that is usually proven in class, and then the proof is often forgotten.
1. The expression contain {{{n+1}}} terms,
which is easy to see/count in the formula above,
because you see {{{1}}} first {{{x^n}}} term without a {{{y}}} , plus
{{{n}}} terms showing {{{y}}} with {{{n}}} different numbers as exponents,
starting with the invisible exponent {{{1}}} in the term {{{n*x^(n-1)y=n*x^(n-1)y^1}}} ,
and going all the way to exponent {{{n}}} in the {{{y^n}}} last term.
2. The exponent of x in the first term is {{{n}}} because the term that we write first is {{{x^n}}} .
There is no compelling reason to write the terms in that order,
but it is customary, and having a certain order in mind helps keep track of all those terms.
3. Since the first term is {{{x^n=x^n*1=x^n*y^0}}} , we can say that the exponent of y in the first term is {{{0}}} . 
4. The sum of the exponents in any term of the expansion is {{{n}}} .
That is true for the first term, {{{x^n=x^n*y^0}}} , and is also true for all of the other terms.


EXPLANATION OF THE FORMULA (in case you care):
The formula comes from the fact that
{{{(x+y)^n=(x+y)*(x+y)*"..."*(x+y)*(x+y)}}} with {{{n}}} factors.
Before simplifying, the product of those {{{n}}} factors would have {{{2^n}}} products made by choosing one of the variables (x or y) from each of the {{{(x+y)}}} factors.
Because each of those {{{2^n}}} products has {{{n}}} factors, the degree of each product (meaning the sum of the exponents of x and y) is {{{n}}} .
Choosing the x from each of the {{{n}}} {{{(x+y)}}} factors, we would get {{{x^n}}} . The only one way to get the product {{{x^n}}} is to choose the {{{x}}} term from all of the {{{n}}} {{{(x+y)}}} factors,
so you get that product only once.
The same can be said of {{{y^n}}} .
Because we like to put x's before y's, and the {{{(x+y)^n}}} already had the x before the y,
{{{x^n}}} is written as the first term, and {{{y^n}}} is written as the last term.
We get other products multiple times
If you choose the y from one of the {{{(x+y)}}} factors and the x from the others, you get products like
{{{y*x*x*x*"..."x*x}}} , {{{x*y*x*x*"..."x*x}}} that can be written as {{{x^(n-1)*y}}} .
Of course there are {{{n}}} different ways to do that, and after simplifying all those {{{n}}} products would be accounted for in the term {{{n*x^(n-1)*y}}} .
There are {{{n(n-1)/2}}} ways to choose {{{2}}} y's and {{{n-2}}} x's,
and there are also {{{n(n-1)/2}}} ways to choose {{{2}}} x's and {{{n-2}}} y's,
and that explains the coefficients in the terms {{{(n(n-1)/2)x^(n-2)y^2}}} and {{{(n(n-1)/2)x^2y^(n-2)}}} .