We will prove the binomial theorem using mathematical induction; that is,
\begin{equation*}
(a + b)^n = \sum_{k = 0}^{n} \binom{n}{k} a^k b^{n - k}\text{,}
\end{equation*}
where \(a\) and \(b\) are real numbers, \(n \in \mathbb{N}\text{,}\) and
\begin{equation*}
\binom{n}{k} = \frac{n!}{k! (n - k)!}
\end{equation*}
is the binomial coefficient. We first show that
\begin{equation*}
\binom{n + 1}{k} = \binom{n}{k} + \binom{n}{k - 1}\text{.}
\end{equation*}
This result follows from
\begin{align*}
\binom{n}{k} + \binom{n}{k - 1} & = \frac{n!}{k!(n - k)!} +\frac{n!}{(k-1)!(n - k + 1)!}\\
& = \frac{(n + 1)!}{k!(n + 1 - k)!}\\
& =\binom{n + 1}{k}\text{.}
\end{align*}
If \(n = 1\text{,}\) the binomial theorem is easy to verify. Now assume that the result is true for \(n\) greater than or equal to 1. Then
\begin{align*}
(a + b)^{n + 1} & = (a + b)(a + b)^n\\
& = (a + b) \left( \sum_{k = 0}^{n} \binom{n}{k} a^k b^{n - k}\right)\\
& = \sum_{k = 0}^{n} \binom{n}{k} a^{k + 1} b^{n - k} + \sum_{k = 0}^{n} \binom{n}{k} a^k b^{n + 1 - k}\\
& = a^{n + 1} + \sum_{k = 1}^{n} \binom{n}{k - 1} a^{k} b^{n + 1 - k} + \sum_{k = 1}^{n} \binom{n}{k} a^k b^{n + 1 - k} + b^{n + 1}\\
& = a^{n + 1} + \sum_{k = 1}^{n} \left[ \binom{n}{k - 1} + \binom{n}{k} \right]a^k b^{n + 1 - k} + b^{n + 1}\\
& = \sum_{k = 0}^{n + 1} \binom{n + 1}{k} a^k b^{n + 1- k}\text{.}
\end{align*}
