Prove the statement by the Principle of Mathematical Induction: For any natural number $n$,$7^{n}-2^{n}$ is divisible by $5$.

(N/A) Let $P(n): 7^{n}-2^{n}$ be divisible by $5$.
For $n=1$:
$P(1): 7^{1}-2^{1} = 7-2 = 5$,which is divisible by $5$.
Therefore,$P(1)$ is true.
Assume that $P(k)$ is true for some $k \in N$:
$P(k): 7^{k}-2^{k} = 5m$,where $m \in N$ (Equation $i$).
We need to prove that $P(k+1)$ is true:
$P(k+1): 7^{k+1}-2^{k+1} = 7 \cdot 7^{k} - 2 \cdot 2^{k}$
$= 7 \cdot 7^{k} - 7 \cdot 2^{k} + 7 \cdot 2^{k} - 2 \cdot 2^{k}$
$= 7(7^{k}-2^{k}) + 2^{k}(7-2)$
$= 7(5m) + 2^{k}(5)$
$= 5(7m + 2^{k})$
Since $5(7m + 2^{k})$ is a multiple of $5$,$P(k+1)$ is true.
By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.