Prove that $21^{n}$ cannot end in zero for any natural number $n \in N$.

(N/A) For a number to end in the digit $0$,it must be divisible by $10$.
This implies that the prime factorization of the number must contain at least one pair of the prime factors $2$ and $5$.
The prime factorization of $21$ is $3 \times 7$.
Therefore,the prime factorization of $21^{n}$ is $(3 \times 7)^{n} = 3^{n} \times 7^{n}$.
Since the prime factors of $21^{n}$ are only $3$ and $7$,it does not contain the prime factor $2$ or $5$.
Because the prime factor $2$ and $5$ are missing,$21^{n}$ is not divisible by $10$.
Hence,$21^{n}$ cannot end in the digit $0$ for any natural number $n \in N$.