This is a presentation of the proof for the binomial formula for complex numbers. The intent is to provide a clear example of an inductive proof.
Let \(z_1, z_2 \in \mathbb C\) and \(n \in \mathbb N\). Then
\[(z_1 + z_2)^n = \sum_{k=0}^n \binom{n}{k}z_1^k z_2^{n-k}\]
By induction on \(n\).
\[\begin{aligned} \sum_{k = 0}^1 \binom{1}{k} z_1^k z_2^{1 - k} &\stackrel{?}{=} (z_1 + z_2)^1\\ \underbrace{\binom{1}{0} z_1^0 z_2^{1-0}}_{k = 0} + \underbrace{\binom{1}{1} z_1^1 z_2^{1 - 1}}_{k = 1} &\stackrel{?}{=} z_1 + z_2\\ 1 \cdot 1 \cdot z_2 + z \cdot z_1 \cdot 1 &\stackrel{?}{=} z_1 + z_2\\ z_2 + z_1 &\stackrel{\checkmark}{=} z_1 + z_2\\ \end{aligned}\]
Thus the base case holds.
Consider \((z_1 + z_2)^{n + 1}\), we wish to show that \(\displaystyle{(z_1 + z_2)^{n + 1} = \sum_{k = 0}^{n + 1} \binom{n + 1}{k}z_1^k z_2^{n + 1 - k}}\).
\[(z_1 + z_2)^{n + 1} = (z_1 + z_2) \cdot (z_1 + z_2)^n\]
\[(z_1 + z_2)^{n + 1} = (z_1 + z_2) \cdot \left(\sum_{k = 0}^n \binom{n}{k} z_1^k z_2^{n - k}\right)\]
Multiplying \((z_1 + z_2)\) through the sum we get
\[\sum_{k = 0}^n \binom{n}{k} z_1 ^{k + 1} z_2^{n - k} + \sum_{k = 0}^n \binom{n}{k} z_1^k z_2^{n + 1 - k}\]
We reindex the first summation \(k \to j - 1\). Note that \(k = j - 1\) and \( k + 1 = j\).
\[\sum_{j = 1}^{n + 1} \binom{n}{j - 1} z_1^j z_2^{n + 1 - j} + \sum_{k = 0}^n \binom{n}{k} z_1^k z_2^{n + 1 - k}\]
We split off the first term of the last summation and the last term of the first summation.
\[\underbrace{\binom{n}{n} z_1^{n + 1} z_2^0}_{j = n+1} + \sum_{k = 1}^n \left[\binom{n}{k - 1} + \binom{n}{k}\right] z_1^k z_2^{n + 1 - k} + \underbrace{\binom{n}{0} z_1^0 z_2^{n + 1}}_{k = 0}\]
Note that \(\binom{n}{k - 1} + \binom{n}{k} = \binom{n + 1}{k}\). This is proved as a lemma later. We then have
\[1 \cdot z^{n + 1} \cdot z_2^0 + \sum_{k = 1}^n \binom{n + 1}{k} z_1^k z_2^{n + 1 - k} + 1 \cdot z_1^0 \cdot z_2^{n + 1}\]
Noting that \(\binom{n}{n} = 1\) and \(\binom{n}{0} = 1\) we have
\[\underbrace{\binom{n + 1}{n + 1} \cdot z^{n + 1} \cdot z_2^0}_{\text{$k = n+1$ term}} + \sum_{k = 1}^n \binom{n + 1}{k} z_1^k z_2^{n + 1 - k} + \underbrace{\binom{n + 1}{0} \cdot z_1^0 \cdot z_2^{n + 1}}_{\text{$k = 0$ term}}\]
We then have the following
\[\sum_{k = 0}^{n + 1} \binom{n + 1}{k}z_1^k z_2^{n + 1 - k}\]
Which is precisely what we wanted to show \((z_1 + z_2)^{n + 1}\) is equal to. Therefore, the statement is true for \(n + 1\) and thus for all \(n\).
\[{\binom{n + 1}{k} = \binom{n}{k} + \binom{n}{k - 1}}\]
Label one of the \(n + 1\) objects as \(*\). We can either choose \(*\) as one of our \(k\) objects or not. These are two distinct and disjoint cases, and are all that are possible.
Therefore, we have \(\displaystyle{\binom{n + 1}{k} = \binom{n}{k} + \binom{n}{k - 1}}\).
Note this lemma could be proved using the definition of \(\displaystyle\binom{n}{k} = \frac{n!}{k!(n - k!)}\), but that would require more algebra than is pleasant. Also note that this proof of the binomial formula for complex numbers exemplifies the three parts of an induction proof:
These three parts together show that the statement is true for all \(n\).
It's also a good idea in proof-writing to be extremely clear both in what proof method you're using and which part of the proof you're currently working on.