If the number \(6^n\) ends with the digit zero. Then it is divisible by 5.
Therefore, the prime factors of \(6^n\) contains the prime number 5. This is not possible because the only primes in the factors of \(6^n\) are 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees that there are no other prime in the factors of \(6^n\).
So, there is no value of n in natural numbers for which \(6^n\) ends with the digit zero.