Prove that $A^s * A^t = A^{s+t}$ if $A \in M_n(\Bbb R)$ a matrix and $s,t \in Z $, prove that $A^s * A^t = A^{s+t}$.
It is obvious that this is true, but the correct proof is a little bit complicated. The obvious way to prove the statement is to use mathematical induction keeping the $s=1$ unchanged and prove the statement for each $t$ and then, making the same thing to the $t$. So,
we define $s=1$. For $t=1$ the statement is true ($A^1 * A^1 = A^2)$. Suppose that the statement is true for each $t$. We have to prove that the statement is correct for $t+1$. In detail, $A^1 *  A^t = A^{t+1}$.
Question 1: How to say that the last equation is true?
The only thing that I could do is to say that $A^t = A^{t-1}*A$ due to induction. So, if I unwrap the left part of the equation I will reach to $t+1$ times of $A$. So, the statement is correct.
Then I define the $t=1$ and suppose that for $s=1$ the statement is correct. I assume that for each $s$ the statement is correct and prove that the statement is correct for $s+1$.
Question 2: Is this the correct way of proving the statement?
Question 3: Is the proof different if the $s$ and/or $t$ are negative numbers?
 A: You might have a more direct route if you show both
\begin{align*}
[t > 0]: \quad A^s A^t &= A^s (A A^{t-1}) = (A^s A) A^{t-1} = A^{s+1} A^{t-1}  \text{, and }  \\
[t < 0]: \quad A^s A^t &= (A^{s-1} A) A^t = A^{s-1} (A A^t) = A^{s-1} A^{t+1}
\end{align*}
so you can organize your induction only along $t$, taking $t$ to zero by single steps.  As you suspect, which of the two lines you use in the display depends on the sign of $t$.
You probably want to carry some version of the statement "the sum of the two new powers is the same as the sum of the two old powers" along your induction.  For instance, at the top of your proof, "let $S = s+t$", and observe that after shuffling one copy of $A$ between the two powers, that $S = (s+1)+(t-1)$ or $S = (s-1)+(t+1)$, depending on which branch of the proof (i.e., which sign of $t$) you are on, so the sum of the powers is constant throughout this induction.
A: Matrix multiplication is associative.
$$(\underbrace{AA\cdots A}_s)(\underbrace{AA\cdots A}_t)=\underbrace{AA\cdots A}_{s+t}.$$

If you define negative exponents to denote powers of the inverse,
$$s<t\to(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_s)(\underbrace{AA\cdots A}_t)=\underbrace{AA\cdots A}_{t-s}.$$
$$s=t\to(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_s)(\underbrace{AA\cdots A}_t)=I=A^0.$$
$$s>t\to(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_s)(\underbrace{AA\cdots A}_t)=\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_{s-t}.$$
