identity of Euler's totient function

Question

Let $p$ be a prime and $\varphi(n)$ be Euler's totient function and $m,n$ be natural numbers. Then $\varphi(p^k)=\frac{p^{k+1}-1}{p-1}$ and $\varphi(mn)=\varphi(m)(n)$ and $\varphi(p)=p-1$.But this $$ \varphi(p^k)=\varphi(p)^k=(p-1)^k\neq \frac{p^{k+1}-1}{p-1}=\varphi(p^k) $$ bothers me. Where is the mistake here?

Answer 1

$\sigma(mn)$ is a multiplicative function, which means that $\sigma(mn)=\sigma(m)\sigma(n)$ when $m$ and $n$ are relatively prime.

A function $f$ is totally multiplicative when $f(mn)=f(m)f(n)$ for all $m$ and $n$. As shown by your example, $\sigma$ is not totally multiplicative.