Show that number of elements in $(Z/nZ)^*$ is $\phi (n)$

Result:"An integer $k\in \Bbb Z/n\Bbb Z$ is a generator of $\Bbb Z/n\Bbb Z$ iff $gcd(n,k)=1$"

Now if $k $ is an integer satisfying $gcd(n,k)=1$ then $k\in (\Bbb Z/n\Bbb Z)^*$

Also as $k$ is a generator of $\Bbb Z/n\Bbb Z$$\Rightarrow$ $|k|=n$

Result:"If $l$ is a positive divisor of $n$, the number of elements of order $l$ in a cyclic group of order $n$ is $\phi(l)$ "

Here in our case $k$ is an element of order $n$. Here $n$ is a divisor of itself (i.e. it satiesfies the hypothesis of the above result) so the number of elements of order $n$ (Which are precisely all of our $k's$) is $\phi(n)$

Thus $(\Bbb Z/n\Bbb Z)^*$ has exactly $\phi(n)$ elements.

The above results are pretty much standard and are taught in almost every basic course in group theory.