A number ending in the digit \(0\) is divisible by \(10\), which implies it's also divisible by \(2\) and \(5\) since \(10 = 2 × 5\).
The prime factorization of \(6^n\) is \((2 × 3)^n\).
Observe that \(5\) is not a prime factor of \(6^n\).
Consequently, \(6^n\) is never divisible by \(5\) for any value of \(n\).
Therefore, \(6^n\) cannot end with the digit \(0\) for any natural number \(n\).