By induction:
Basic Step: n = 1 is trivial
Inductive Step: Assume the statement is true as given for some \(n \ge 1\text{.}\) Show \((a+b)^{n+1} = \sum_{k=0}^{n+1} {\binom{n+1}{k} a^k b^{n+1-k}}\)
\begin{align*} (a+b)^{n+1} & = (a+b)(a+b)^n\\ & = (a+b)\sum_{k=0}^{n} {\binom{n}{k} a^k b^{n-k}}\\ & = \sum_{k=0}^n \binom{n}{k} a^{k+1} b^{n-k} + \sum_{k=0}^n \binom{n}{k} a^k b^{n-k+1}\\ & = \sum_{k=0}^{n-1} \binom{n}{k} a^{k+1} b^{n-k} + a^{n+1} + b^{n+1} + \sum_{k=1}^n \binom{n}{k} a^k b^{n-k+1}\\ & = \sum_{j=1}^n \binom{n}{j-1} a^j b^{n-(j-1)} + a^{n+1} + b^{n+1} + \sum_{k=1}^n \binom{n}{k} a^k b^{n+1-k}\\ & = b^{n+1} + \sum_{k=1}^n \left [ \binom{n}{k-1} + \binom{n}{k} \right ] a^k b^{n+1-k} + a^{n+1}\\ & = b^{n+1} + \sum_{k=1}^n \binom{n+1}{k} a^k b^{n+1-k} + a^{n+1}\\ & = \sum_{k=0}^{n+1} \binom{n+1 }{k} a^k b^{n+1-k} \end{align*}