For $b\in\mathbb{Z}$ and $k_{1},k_{2}\in\mathbb{Z}_{\ge0}$, we have the following rules for exponents:
i) $b^{k_{1}}b^{k_{2}} = b^{k_{1}+k_{2}}$
ii)$(b^{k_{1}})^{k_{2}}=b^{k_{1}k_{2}}$
A. Deduce those laws for elements $\alpha\in\mathbb{Z}/m\mathbb{Z}$ and $k_{1},k_{2}\in\mathbb{Z}_{\ge0}$
B. Moreover if $\alpha\in\mathbb{Z}/m\mathbb{Z}$ and $\alpha^*\in\mathbb{Z}/m\mathbb{Z}$ with $\alpha\alpha^* = 1 $ in $\mathbb{Z}/m\mathbb{Z}$, for $k\in\mathbb{Z}_{\ge0}$ we define $\alpha^{-k}:=(\alpha^*)^k$. Show that these two rules can be shown to hold for all $k_{1},k_{2}\in\mathbb{Z}$ (by converting everything to use positive exponents)
$\it{Hint}:$ if such $\alpha,\alpha^*\in\mathbb{Z}/m\mathbb{Z}$ exist, then show that $(\alpha^*)^*=\alpha$ and consider four cases for each of i) and ii) depending on the signs of $k_{1},k_{2}$
This is what I have so far for part A
$\alpha^{k_{1}}\alpha^{k_{2}} = ([a+b]_{m})^{k_{1}}([a+b]_{m})^{k_{2}})=([a+b]_{m})^{k_{1}+k_{2}}$
$(\alpha^{k_{1}})^{k_{2}} = ([a+b]_{m})^{k_{1}})^{k_{2}}=([a+b]_{m})^{k_{1}k_{2}}$
Is this correct?
Hints:
Part A:
$\alpha = x \pmod m$, this means that $x=mq+\alpha$. They ask you to prove that $\alpha^{k_1}\alpha^{k_2}=\alpha^{k_1+k_2}$, and this follows from $(mq+\alpha)^{k_1}(mq+\alpha)^{k_2}$, because $mq+\alpha\in \mathbb{Z}$ is $(mq+\alpha)^{k_1}(mq+\alpha)^{k_2}= (mq+\alpha)^{k_1+k_2}$. With a similar reasoning you prove the second rule.
Part B:
You need to consider the signs of $k_1, k_2$ in the four possibilities.
If $k_1, k_2>0$ then we have it proved by means of part A.
If $k_1, k_2 <0$ then $k_1+k_2<0$, this means $\alpha^{k_1+k_2}=(\alpha^{-k_2-k_1})^*$. $-k_2-k_1>0$, then by the first rule $(\alpha^{-k_2-k_1})^*=(\alpha^{-k_2}\alpha^{-k_1})^*$ and we conclude $(\alpha^{-k_1})^*(\alpha^{-k_2})^*=\alpha^{k_1}\alpha^{k_2}$.
If $k_1>0, k_2 <0$, then $k_1+k_2>0$ or $k_1+k_2<0$. First case: consider $\alpha^{k_1+k_2}\alpha^{-k_2}\alpha^{-k_1}$, both $k_1+k_2$ and $-k_2$ are greater than zero, then $\alpha^{k_1+k_2}\alpha^{-k_2}\alpha^{-k_1}=\alpha^{k_1+k_2-k_2}\alpha^{-k_1}=\alpha^{k_1}(\alpha^{k_1})^*=1$. This means that $\alpha^{k_1+k_2}=(\alpha^{-k_2}\alpha^{-k_1})^*=\alpha^{k_1}\alpha^{k_2}$. You finish this part proving the same when $k_1+k_2<0$.
Similarly you prove the second rule. For example, if $k_1>0, k_2 <0$ we have $(\alpha^{k_1})^{k_2}=((\alpha^{k_1})^{-k_2})^*=(\alpha^{k_1(-k_2)})^*=(\alpha^{-k_1k_2})^*=\alpha^{k_1k_2}$.