If $G=\left<a\right>$ is cyclic of order $n$, then $a^k$ is also a generator iff $(k,n)=1$. Conclude that the number of generators of $G$ is $\phi(n)$.
I think it's easy but I'm just stuck in both directions in the first part. I've tried to use Bezout's Lemma and the division algorithm but getting nowhere.
Can someone give some hint? Thanks in advance
Bezout's Lemma is the right direction. $(k,n)=1$ means that there exist integers $x,y$ such that $xk+yn=1$. Then you have $$(a^k)^x=a^{kx}=a^{1-yn}=a(a^{-n})^y=a(1^y)=a.$$ Since you can obtain $a$ from a power of $a^k$, you can obtain any $a^z$ similarly via $(a^k)^{xz}=a^z$. Totient just counts the number of distinct $k$ such that $(k,n)=1$.
The basic fact that you need to prove it is: $|a^k|=n/(n,k)$.
This is straight forward and a good exercise to prove.
So, first, it's easy to see that $n|kn/(n,k)$, so that $(a^k)^{n/(n,k)}=e$.
Next, any $d$ such that $(a^k)^d=e$ satisfies $n|kd$. This implies $kd\ge\operatorname{lcm}(n,k)$. But if $d\lt n/(n,k)$ then $kd\lt \operatorname{lcm}(n,k)\Rightarrow\Leftarrow$.