Note that by the binomial theorem $$4^n=(3+1)^n=3^n+\sum_{k=1}^{n-1}\binom{n}{k}\cdot 3^k\cdot 1^{n-k}+1^n.$$ Hence
$$4^n-3^n-1=\sum_{k=1}^{n-1}\binom{n}{k}\cdot 3^k.$$ Show that for $n=11$ (any prime will work) then the binomial coefficients $\binom{11}{k}$ for $k=1,\dots,10$ are all multiple of $11$.