I am beginning to learn how to write proofs and I would like some verification on this simple proof I have done for the sum of the interior angles of a polygon. I thought this would be a good one to prove since it is pretty basic. So, I have the following hypothesized functional relation for the angle measures of a polygon being $$t(n+1)=t(n) + \pi$$ where $t(n)=(n-2)\pi$. I want to show that this is true for all values of $n\ge2$ in the natural numbers. If I have any incorrect wording or baseless assumptions or steps missing, I would appreciate it if you could point them out.
Proof:
Base Case: $t(3)=(3-2)\pi = \pi =t(2)+\pi=0+\pi=\pi$.
Induction Step: Assume $t(n+1)=t(n) + \pi$ is true for any arbitrary value $n=k\ge2 \, \forall \, n,k \in \mathbb{N}$. Then $$\begin{align*} t(k+1) &= t(k)+\pi \\ &= (k-2)\pi+\pi \end{align*}$$
and $$\begin{align*} t(k+2) &= (k+2-2)\pi\\ & = ([k+1]-2+1)\pi \\ &= ([k+1]-2)\pi+\pi \\ &=t(k+1)+\pi \end{align*} $$
Sorry if this is too easy or trivial to be on here, just want some clarification on my wording and format. Also, is this the same as proving that the total interior angle measure of a polygon is equal to $(n-2)\pi$ for all natural numbers greater than or equal to 2?
Edit: I believe in the first part of the induction step, I go in reverse when showing the expansion of $t(k+1)$. It should start as $t(k+1)=(k+1-2)\pi$. Also, is that step redundant? I could have just proved it from that step initially. I believe I am also missing the step where I assume $t(n)$ to be true for all specified values.
Best Answer
This might provide some additional clarity:
In the "inductive step", you may assume that $t\left(k\right)$ is true, and then show that $t\left(k+1\right)=t\left(k\right)+\pi$:
$$ t\left(k+1\right)=\left(\left(k+1\right)-2\right)\pi $$
from the definition of $t\left(n\right)$. This leads to:
$$ t\left(k+1\right)=\left(\left(k-2\right)+1\right)\pi $$
$$ t\left(k+1\right)=\left(k-2\right)\pi+\pi $$
$$ t\left(k+1\right)=t\left(k\right)+\pi $$
I hope this helps.