The equation is true for $n=1,2,3,4$. Notice that $$s(ab)\leq s(a)s(b)$$for any $a,b$ positive integers. Therefore, $$s(11^n)\leq s(11^{n-1})s(11)=2s(11^{n-1})\leq 2^2s(11^{n-2})\leq \cdots\leq 2^{n-1}s(11)=2^{n}$$So, if $s(11^{n_0})=2^{n_0}$ for some positive integer $n_0$, then $s(11^n)=2^n$ for any positive integer $n$ less than $n_0$. Because $s(11^5)\neq 2^5$, there is no solution for $n\geq 5$. Only solutions are $\boxed{n=1,2,3,4}$

By the Binomial Theorem, the coefficients of $11^n$ add to $2^n$. Therefore, as long as there is no carry-over, $s(11^n) = 2^n$, meaning we have $n = 1,2,3,4 \Rightarrow \boxed{10}$.