In order to check whether 6^{n} can end with the digit 0 for any natural number n, let us find the factors of 6.

It’s seen that the factors of 6 are 2 and 3.

So, 6^{n} = (2 x 3)^{n}

6^{n} =2^{n} x 3^{n}

Since, the prime factorization of 6 does not contain 5 and 2 as its factor, together. We can thus conclude that 6^{n} can never end with the digit 0 for any natural number n.