Statement

Proof

For $1 \leq j < k$, $p_j^{e_j}$ is coprime to $\prod_{i=j+1}^k p_i^{e_i}$. So by repeated application of Claim 1 we have

$$\phi(n) = \prod_{i=1}^k \phi(p_i^{e_i}).$$

Which from Claim 2 we get the result

$$\phi(n) = \prod_{i=1}^k (p_i - 1)p_i^{e_i-1} = n \cdot \prod_{i=1}^k \left ( 1 - \frac{1}{p} \right ).$$

Claim 1

Proof of Claim 1

As $m$ and $n$ are coprime the Chinese remainder theorem gives has a bijection

$$b: \{i \vert 0 \leq i < m\} \times \{i \vert 0 \leq i < n\} \rightarrow \{i \vert 0 \leq i < mn\}$$

where $b(x,y) = x$ (mod $m$) and $b(x,y) = y$ (mod $n$).

Note if $gcd(b(x,y),mn) = 1$ if and only if $gcd(x,m) = 1$ and $gcd(y,n) = 1$ as $m$ and $n$ are coprime. Therefore

$$b: \begin{array}\ \{ x \in \mathbb{N} \ \vert \ 0 < x \leq m, \ gcd(x,m) = 1 \} \times \\ \ \{y \in \mathbb{N} \ \vert \ 0 < x \leq n, \ gcd(x,n) = 1 \} \end{array} \mapsto \left \{b(x,y) \in \mathbb{N} \ \Bigg \vert \begin{array} \ 0 < b(x,y) \leq mn, \\ \ gcd(b(x,y),mn) = 1 \end{array} \right \}$$

and we have

$$\phi(mn) = \phi(m) \phi(n).$$

Claim 2

Proof of Claim 2

As $p$ is prime we know the positive divisors of $p^k$ are powers of $p^i$ for $0 \leq i \leq k$.

Therefore of $gcd(p^k, x) > 1$ we know $x$ will have to be a multiple of $p$.

So $x = mp$ with $m \in \mathbb{Z}_{\not = 0}$ and for $0 < x < p^k$ we have $0 < m \leq p^{k-1}$.

So

$$\phi(p^k) = \big \vert \left \{x \in \mathbb{N} \ \vert \ 0 < x \leq p^k, \ gcd(x,p^k) = 1 \right \} \ \big \vert = p^k - p^{k-1}$$

giving the desired result.