Power of Identity is Identity

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity element is $e$.

Then:


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

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

Basis for the Induction
By definition of power of group 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:

Therefore, by Principle of Mathematical Induction:


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

Now we prove the case $n < 0$.

We have:

Thus:


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