An elaboration of Thm. 8.11(c) proof. The theorem is given below$ should be apparent but my proof to it was as follows: Since the index definition is given by $r^{\operatorname{ind} a} \equiv a \pmod n$, then upon substituting $a = 1$ in this formula we get $r^{\operatorname{ind} 1} \equiv 1 \pmod n$, which means that $\operatorname{ind} 1 = 0 \pmod n $ but I need it to be $\operatorname{ind} 1 = 0 \pmod {\phi(n)} $ instead, could anyone explain this for me please?

The (sub)group of units in $\mathbb{Z}/n\mathbb{Z}$ is the group $\left(\mathbb{Z}/n\mathbb{Z}\right)^\times=\\{\overline{a}:\gcd(a,n)=1\\}$. This group has $\phi(n)$ elements.

A number $r$ is called a primitive root if $\overline{r}$ is a generator of this group $\left(\mathbb{Z}/n\mathbb{Z}\right)^\times$, this is the same as saying that the order of $r$ mod $n$ is $\phi(n)$ (i.e. if $r^{m}\equiv 1\pmod n$, then $\phi(n)|m$).

So, when you have $r^{\operatorname{ind} 1}\equiv 1\bmod n$, this implies that $\phi(n)|\operatorname{ind} 1$.