Powers of 16 Modulo 20/Proof 1

Proof
Proof by induction:

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
 * $16^n \equiv 16 \pmod {20}$

Basis for the Induction
$\map P 1$ is the case:
 * $16^1 \equiv 16 \pmod {20}$

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $16^k \equiv 16 \pmod {20}$

from which it is to be shown that:
 * $16^{k + 1} \equiv 16 \pmod {20}$

Induction Step
This is the induction step:

We have:

So $\map P k \implies \map P {k + 1}$ and then it follows by the Principle of Mathematical Induction that:


 * $\forall n \in \Z_{\ge 0}: 16^n \equiv 16 \pmod {20}$