6 is 1+5, so the binomial expansion of 6^n=(1+5)^n=1+5n+25n(n-1)/2!+125n(n-1)(n-2)/3!+... . This can be written:
1+5n+25K, where K is an integer, because 25 is a factor in all successive terms. So 6^n=1+5n+25K. Rearranging, we get: 6^n-5n=25K+1. Therefore, if 25 is divided into the right-hand side there will always be a remainder of 1, so it follows that 6^n-5n will also give a remainder of 1 when divided by 25, because of the equality.