Binomial Theorem: $\sum\limits_{k=0}^{n}{n \choose k}x^ky^{n-k} = (x+y)^n$

$\sum\limits_{k=0}^{n}{n \choose k}x^ky^{n-k} = (x+y)^n$ since:

(1)  For $n=0, \sum\limits_{k=0}^{n}{n \choose k}x^ky^{n-k} = {0 \choose 0}x^{0}y^{0} = 1 = (x+y)^0$

(2)   Assume that $\sum\limits_{k=0}^{n}{n \choose k}x^ky^{n-k} = (x+y)^n$ 

(3)  $(x+y)(x+y)^n =$ $$= (x+y)(x^n + {n \choose 1}x^{n-1}y + {n \choose 2}x^{n-2}y^2 + \dots + {n \choose k}x^{n-k}y^k +$$ $$+ \dots {n \choose {n-1}}xy^{n-1} + y^n)$$  

$$= x^{n+1} + \left[1 + {n \choose 1}\right]x^ny + \left[{n \choose 1} + {n \choose 2}\right]x^{n-1}y^2 + \dots$$ $$\left[{n \choose k-1} + {n \choose k}\right]x^{n-k+1}y^k + \dots + \left[{n \choose {n-1}} + 1\right]xy^n + y^{n+1}$$

(4) Using Pascal's Rule ${n \choose {r-1}} + {n \choose r} = {{n+1} \choose r}$ and $1 = {n \choose 0} = {n \choose n}$ so that:

$$x^{n+1} + \left[1 + {n \choose 1}\right]x^ny + \left[{n \choose 1} + {n \choose 2}\right]x^{n-1}y^2 + \dots$$ $$\left[{n \choose k-1} + {n \choose k}\right]x^{n-k+1}y^k + \dots + \left[{n \choose {n-1}} + 1\right]xy^n + y^{n+1} =$$

$${{n+1} \choose 0}x^{n+1} + {{n+1} \choose 1}x^ny + {{n+1} \choose 2}x^{n-1}y^2 + \dots$$ $${{n+1} \choose k}x^{n-k+1}y^k + \dots + {{n+1} \choose {n-1}}xy^n + {{n+1}\choose {n+1}}y^{n+1} = \sum\limits_{k=0}^{n+1}{{n+1} \choose k}x^ky^{n-k}$$