Let \(n = p_1 ^ {a_1} p_2 ^ {a_2} \cdots p_k ^ {a_k}\), where \(p_i\) is a prime factor of \(n\) and \(a_1 \ge 1\). The formula for \(\phi(n)\) (derived here) is:
\[\begin{align} \phi(n) &= n \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right) \\ &= \prod_{i=1}^k p_i^{a_i} \left( 1 - \frac{1}{p_i} \right) \end{align}\]
If we raise \(n\) to the power of \(j\):
\[\begin{align} n^j &= (p_1 ^ {a_1} p_2 ^ {a_2} \cdots p_k ^ {a_k})^j \\ &= p_1 ^ {ja_1} p_2 ^ {ja_2} \cdots p_k ^ {ja_k} \end{align}\]
If we use the Euler phi function with \(n^j\):
\[\phi(n^j) = \prod_{i=1}^k p_i^{ja_i} \left( 1 - \frac{1}{p_i} \right) \]
Multiplying with \(p_i^{a_i} / p_i^{a_i}\):
\[\begin{align} \phi(n^j) &= \prod_{i=1}^k \frac{p_i^{a_i}}{p_i^{a_i}} p_i^{ja_i} \left( 1 - \frac{1}{p_i} \right) \\ \phi(n^j) &= \prod_{i=1}^k p_i^{a_i} \frac{p_i^{ja_i}}{p_i^{a_i}} \left( 1 - \frac{1}{p_i} \right) \end{align}\]
Taking \(p_i^{ja_i} / p_i^{a_i}\) out:
\[\begin{align} \phi(n^j) &= \frac{p_1^{ja_1}p_2^{ja_2} \cdots p_k^{ja_k}}{p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}} \prod_{i=1}^k p_i^{a_i} \left( 1 - \frac{1}{p_i} \right) \\ &= \frac{n^j}{n} \prod_{i=1}^k p_i^{a_i} \left( 1 - \frac{1}{p_i} \right) \end{align}\]
This means:
\[ \phi(n^j) = n^{j-1} \phi(n)\]