Peter wrote: Since for $n=1$ there stands $25$, it can be at most $25$, we will show it is $25$. Since $16^{n}$ ends in $16, 56, 96, 36, 76$, taking the remainder mod $25$ gives $16,6,21,11,1$, substract $10n\bmod 25$ from that and you have $1,1,1,1,1,\ldots$, so every element is a multiple of $25$.

Another way: $ 16^{n}+10 n-1\equiv (1+15)^{n}-15n-1\equiv\sum_{k = 2}^{n}{n\choose k}15^{k}\equiv 0\bmod 5^{2}$, for any $ n = 1, 2,\ldots$ Since $ 16^{1}+15\cdot 1-1 = 5^{2}$, the greatest common divisor we're looking for is exactly $ 5^{2}$.

Another way, let f(n)=16^n+10n-1. Then 16f(n)=16^(n+1)+160n-16=16^(n+1)+150n+(10(n+1)-1)-(16+10-1)=f(n+1)+150n-25. Thus if f(n) is divisible by 25, so is f(n+1).