Product of Complex Conjugates/General Result

Theorem
Let $z_1, z_2, \ldots, z_n \in \C$ be complex numbers.

Let $\overline z$ be the complex conjugate of the complex number $z$.

Then:
 * $\displaystyle \overline {\prod_{j \mathop = 1}^n z_j} = \prod_{j \mathop = 1}^n \overline {z_j}$

That is: the conjugate of the product equals the product of the conjugates.

Proof
Proof by induction:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \overline {\prod_{j \mathop = 1}^n z_j} = \prod_{j \mathop = 1}^n \overline {z_j}$

$P \left({1}\right)$ is trivially true, as this just says $\overline {z_1} = \overline {z_1}$.

Basis for the Induction
$P \left({2}\right)$ is the case:
 * $\overline {z_1 z_2} = \overline {z_1} \cdot \overline {z_2}$

which has been proved in Product of Complex Conjugates.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \overline {\prod_{j \mathop = 1}^k z_j} = \prod_{j \mathop = 1}^k \overline {z_j}$

Then we need to show:
 * $\displaystyle \overline {\prod_{j \mathop = 1}^{k+1} z_j} = \prod_{j \mathop = 1}^{k+1} \overline {z_j}$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \N: \overline {\prod_{j \mathop = 1}^n z_j} = \prod_{j \mathop = 1}^n \overline {z_j}$