Any positive integer ending with the digit zero is divisible by 5 and so its prime factorisations must contain the prime 5.

6^{n} = (2 × 3)^{n} = 2^{n} × 3^{n}

=> The prime in the factorisation of 6^{n} is 2 and 3.

=> 5 does not occur in the prime factorisation of 6^{n} for any n.

=> 6^{n} does not end with the digit zero for any natural number n.