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Using binomial theorem, prove that 6^n-5...

Using binomial theorem, prove that `6^n-5n`always leaves remainder 1 when divided by 25.

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`6^n = (1+5)^n`
`= C(n,0)1^n5^0+C(n,1)1^(n-1)5^1+C(n,2)1^(n-2)5^2+C(n,3)1^(n-3)5^3+...C(n,n)1^0 5^n`
`6^n=1+5n+25(C(n,2)+C(n,3)5+C(n,4)5^2+...C(n,n)5^(n-2))`
`6^n-5n = 1+25k`, where ` k = C(n,2)+C(n,3)5+C(n,4)5^2+...C(n,n)5^(n-2)`
As, `k` is always positive, so `25k` will always be divisible by `25`.
Thus, if we divide `6^n-5n` by `25`, remainder will be `1`.
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