Show that $G:=\mathbb{Z}_{13}^*$ is cyclic I need to prove that $G:=\mathbb{Z}_{13}^*$ (without zero with multipcation)is cyclic 
My attempt:
I tried to check each element in $G$ if it is a generator or not:
$$
\begin{align}
&1^1=1\mod 13\\
&1^2=1\mod 13\\
&1^3=1\mod 13\\
&...\\
&2^1=2\mod 13\\
&2^2=4\mod 13\\
&2^3=8 \mod 13\\
&2^4=3 \mod 13\\
&2^5=6 \mod 13\\
&2^6=12 \mod 13\\
&2^7=11 \mod 13\\
&2^8=9 \mod 13\\
&2^9=5 \mod 13\\
&2^{10}=10 \mod 13\\
&2^{11}=7 \mod 13\\
&2^{12}=1 \mod 13\\
\end{align}
$$
$\Rightarrow$ one is not generator, two is a generator, is it correct so far? is there other way check this?
 A: Very nice, you have found a generator for the group, hence it is cyclic.
Another proof is as follows: $\mathbb Z_{13}$ is a finite field, the multiplicative group of a finite field is always cyclic.
This can be proven in various way, Here is a sketch of one such proof:
Let $n$ be the number of elements of the multiplicative group of the field,let $m$ be the least common multiple of the orders of the elements of the group,then the polynomial $x^m-1$ has $n$ roots since every element satisfies $x^m=1$. On the other hand a polynomial of degree $m$ in a field has at most $m$ roots. So $n\leq m$. On the other hand the degree of every element in $G$ is a divisor of $n$. so $m\leq n$. From here $m=n$.
Apply the fundamental theorem of finite abelian groups, the group is a direct product of cyclic groups of prime power order, if the least common multiple of the orders is equal to the product it is because there is at most one factor for each prime, in other words the group is cyclic.
A: Yes, this is the best way to check this from the tools you have at your disposal (I'm guessing).  You will soon come across ways to find all the generators.
