Example 1 Solve the congruence equation $x^{8} \equiv 41(\bmod 23)$.
From Example 1 in §2, we know that 5 is a primitive root modulo 23. Since \(41 \equiv -5 \pmod{23}\), we find from Table 2 in Example 1 of §3 that \(\gamma_{23,5}(41) = 12\). Therefore, we need to solve the congruence equation:
$$8 y \equiv 12 \pmod{22}$$

Since \((8, 22) = 2 \mid 12\), the above congruence equation has solutions, and the number of solutions is 2. It is easy to see that its two solutions are
$$y \equiv -4, 7 \pmod{22}$$

From Table 1 in Example 1 of §3, we find that the number with index \(18 (18 \equiv -4 \pmod{22})\) is 6; the number with index 7 is -6. Therefore, the complete solutions to the original equation are
$$x \equiv -6, 6 \pmod{23}.$$

When \(m\) has a primitive root \(g\),
$$a = g^j, \quad 0 \leqslant j < \varphi(m)$$

is a reduced residue system modulo \(m\), and \(\gamma(a) = j\). Therefore, by Theorem 1, \(a\) is an \(n\)-th power residue modulo \(m\) if and only if \((n, \varphi(m)) \mid j\). Thus, among the \(\varphi(m)\) numbers given by (6), exactly \(\varphi(m) / (n, \varphi(m))\) numbers are \(n\)-th power residues modulo \(m\):
$$a = g^j, \quad j = t \cdot (n, \varphi(m)), \quad 0 \leqslant t < \varphi(m) / (n, \varphi(m))$$

In this way, we have proved that