Binomial Theorem

Theorem:$$(x+y)^n = \sum_{k=0}^n {n\choose k}x^{n-k}y^k$$, for all $$x,y \in R$$ and all integers$$ n \geq 0$$

Proof:
Base Case:$$n=1$$ $$(x+y)^1=x+y={1\choose 0}x^{1-0}y^0+{1\choose 1}x^{1-1}y^1=\sum_{k=0}^1 {1\choose k}x^{n-k}y^k$$ Therefore the base case holds

Inductive Hypothesis:$$(x+y)^j = \sum_{k=0}^n {j\choose k}x^{j-k}y^k$$ for some $$j \ge 1$$

Inductive Step: $$(x+y)^{n+1}=(x+y)(x+y)^n$$ $$=x\sum_{k=0}^n {n\choose k}x^{n-k}y^k+y\sum_{k=0}^n {n\choose k}x^{n-k}y^k$$ $$=\sum_{k=0}^n {n\choose k}x^{n+1-k}y^k+\sum_{k=0}^n {n\choose k}x^{n-k}y^{k+1}$$ $$={n\choose 0}x^{n+1}+\sum_{k=1}^n {n\choose k}x^{n+1-k}y^k+{n\choose n}y^{n+1}+\sum_{k=0}^{n-1} {n\choose k}x^{n-k}y^{k+1}$$ $$=x^{n+1}+y^{n+1}+\sum_{k=1}^n {n\choose k}x^{n+1-k}y^k+\sum_{k=0}^{n-1} {n\choose k}x^{n-k}y^{k+1}$$ $$={n+1\choose 0}x^{n+1}+{n+1\choose n+1}y^{n+1}+\sum_{k=1}^n {n\choose k}x^{n+1-k}y^k+\sum_{k=1}^n {n\choose k-1}x^{n+1-k}y^k$$ $$={n+1\choose 0}x^{n+1}+{n+1\choose n+1}y^{n+1}+\sum_{k=1}^n\Bigg[{n\choose k}+{n\choose k-1}\Bigg] x^{n+1-k}y^k$$ Then because $${n\choose k}+{n\choose k-1}={n+1\choose k}$$ (see here) we have $$={n+1\choose 0}x^{n+1}+{n+1\choose n+1}y^{n+1}+\sum_{k=1}^n{n+1\choose k}x^{n+1-k}y^k$$ $$=\sum_{k=0}^{n+1}{n+1\choose k}x^{n+1-k}y^k$$ And so we are done by the principle of mathematical induction.