Prove that $6^{n}$ cannot end with the digit $0$ for any natural number $n \in N$.

(N/A) If $6^{n}$ ends with the digit $0$,then it must be divisible by $10$,which means it must be divisible by both $2$ and $5$.
The prime factorization of $6^{n}$ is $(2 \times 3)^{n} = 2^{n} \times 3^{n}$.
According to the Fundamental Theorem of Arithmetic,the prime factorization of any number is unique. The prime factors of $6^{n}$ are only $2$ and $3$.
Since $5$ is not a prime factor of $6^{n}$,$6^{n}$ is not divisible by $5$.
Therefore,$6^{n}$ cannot be divisible by $10$,which implies that $6^{n}$ cannot end with the digit $0$ for any natural number $n \in N$.