# Power of Identity is Identity

## Theorem

Let $\struct {M, \circ}$ be a monoid whose identity element is $e$.

Then:

$\forall n \in \Z: e^n = e$

## Proof

Since $e$ is invertible, the power of $e$ is defined for all $n \in \Z$.

We prove the case $n \ge 0$ by induction.

### Basis for the Induction

By definition of power of monoid element:

$e^0 = e$

so the theorem holds for $n = 0$.

This is our basis for the induction.

### Induction Hypothesis

Our induction hypothesis is that the theorem is true for $n = k$:

$e^k = e$

### Induction Step

In the induction step, we prove that the theorem is true for $n=k+1$.

We have:

 $\displaystyle e^{k + 1}$ $=$ $\displaystyle e^k \circ e$ Definition of Power of Element of Monoid $\displaystyle$ $=$ $\displaystyle e^k$ Definition of Identity Element $\displaystyle$ $=$ $\displaystyle e$ Induction Hypothesis

Therefore, by Principle of Mathematical Induction:

$\forall n \in \Z_{\ge 0} : e^n = e$

$\Box$

Now we prove the case $n < 0$.

We have:

 $\displaystyle e^n$ $=$ $\displaystyle \paren {e^{-n} }^{-1}$ Definition of Power of Element of Monoid $\displaystyle$ $=$ $\displaystyle e^{-1}$ since $-n > 0$ $\displaystyle$ $=$ $\displaystyle e$ Inverse of Identity Element is Itself

Thus:

$\forall n \in \Z : e^n = e$

$\blacksquare$