Proof of the following property about prime numbers [closed]

Is there a proper proof of the following property:

Let $p$ be a prime number. The number of invertible elements in $\mathbb{Z}/p^n\mathbb{Z}$ is $(p-1)p^{n-1}$.

closed as off-topic by user26857, Davide Giraudo, C. Falcon, user228113, ThomasJul 11 '16 at 23:15

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• What makes a proof proper? – vadim123 May 17 '16 at 15:36
• Yes, there is.${}$ – user228113 Jul 11 '16 at 21:49

An element $\overline{a}$ (where $0\leq a \leq n-1)$ of the ring $\mathbb{Z}/n$ is invertible precisely when $a$ is coprime to $n$ by a standard result of elementary number theory. The number of positive integers less than or equal to $n$ which are coprime to $n$ is given by the euler-phi function.
Hence it suffices to compute $\varphi(p^n)$. You can show that $\varphi(p^n)=(p-1)p^{n-1}$ by noting that the only postive integers less than or equal to $p^n$ which are not coprime to it are the multiples of $p$ namely $kp$ for $k=1,\dotsc, p^{n-1}$. Hence $$\varphi(p^n)=p^n-p^{n-1}=(p-1)p^{n-1}.$$
Let $G=\mathbb{Z}/p^{n}\mathbb{Z}$.
Lemma. For $a,n\in\mathbb{N}$, $ax\pmod{m}=1$ has a solution if and only if $\gcd(a,m)=1$.
Proof. $ab\pmod{m}=1\iff m\mid ab-1\iff\exists k\in\mathbb{N}$ such that $mk=ab-1\iff1=a(b)+m(-k)\iff\gcd(a,m)=1$.//
Now, by our lemma we know that the invertible elements in $G$ are precisely those that are relatively prime to $p^{n}$. If you are familiar with the Euler-totient function, then you know that the number of such elements is $\varphi(p^{n})=p^{n}-p^{n-1}=(p-1)p^{n-1}$.