Using the binomial theorem,prove that $6^{n} - 5n$ always leaves a remainder of $1$ when divided by $25$ for all $n \in N$.

For two numbers $a$ and $b$,if we can find numbers $q$ and $r$ such that $a = bq + r$,then we say that $b$ divides $a$ with $q$ as the quotient and $r$ as the remainder. Thus,to show that $6^{n} - 5n$ leaves a remainder of $1$ when divided by $25$,we prove that $6^{n} - 5n = 25k + 1$,where $k$ is some integer.
We have the binomial expansion:
$(1 + a)^{n} = {}^{n}C_{0} + {}^{n}C_{1}a + {}^{n}C_{2}a^{2} + \dots + {}^{n}C_{n}a^{n}$
For $a = 5$,we get:
$(1 + 5)^{n} = {}^{n}C_{0} + {}^{n}C_{1}(5) + {}^{n}C_{2}(5^{2}) + \dots + {}^{n}C_{n}(5^{n})$
$6^{n} = 1 + 5n + 25({}^{n}C_{2} + {}^{n}C_{3}(5) + \dots + 5^{n-2})$
$6^{n} - 5n = 1 + 25({}^{n}C_{2} + 5 \cdot {}^{n}C_{3} + \dots + 5^{n-2})$
Let $k = {}^{n}C_{2} + 5 \cdot {}^{n}C_{3} + \dots + 5^{n-2}$.
Then $6^{n} - 5n = 25k + 1$.
This shows that when $6^{n} - 5n$ is divided by $25$,the remainder is $1$.