Pascal and Binomial Coefficients

I realized that I never really showed how Pascal's Triangle is related to the Binomial Coefficients... So here we go:

(x+1)0 = 
1

(x+1)1 =
1x + 1 =
1 1

(x+1)2 =
(x+1)(x+1) = 
x*x + x*1 + 1*x + 1*1 = 
1x2 + 2x + 1 = 
1 2 1

(x+1)3 =
(x+1)(x+1)(x+1) = 
((x+1)(x+1))(x+1) = 
(1x2 + 2x + 1)(x+1) = 
(1x2*x + 1x2)+(2x*x + 2x*1)+(1*x + 1*1)= 
1x3 + 1x2 + 2x2 + 2x + x + 1 = 
1x3 + 3x2 + 3x + 1 =
1 3 3 1


(x+1)4 =
(x+1)(x+1)(x+1)(x+1) =
((x+1)(x+1)(x+1))(x+1) = 
(1x3 + 3x2 + 3x + 1)(x+1) =
(1x3*x + 1x3*1)+(3x2*x + 3x2*1)+(3x*x + 3x*1)+(1*x + 1*1) =
1x4 + 1x3 + 3x3 + 3x2 + 3x2 + 3x + 1x + 1 =
1x4 + 4x3 + 6x2 + 4x + 1 = 
1 4 6 4 1