11's remainders

$\begin{aligned} S_1 & = \sum_{n=6}^{10} (2n+1)^{n-5} \\ & = \sum_{n=1}^5 (2n+11)^n \\ \implies S_1 & \equiv \sum_{n=1}^5 (2n)^n \pmod {11} \\ & \equiv (2^1+4^2+6^3+8^4+10^5) \pmod {11} \\ & \equiv (2+5+(-5)^3+(-3)^4+(-1)^5) \pmod {11} \\ & \equiv (2+5-4+4-1) \pmod {11} \\ & \equiv 6 \pmod {11} \end{aligned}$

$\begin{aligned} S_2 & = \sum_{n=6}^{10} (2n)^{n-5} \\ & = \sum_{n=1}^5 (2n+10)^n \\ \implies S_2 & \equiv \sum_{n=1}^5 (2n-1)^n \pmod {11} \\ & \equiv (1^1+3^2+5^3+7^4+9^5) \pmod {11} \\ & \equiv (1+9+4+(-4)^4+(-2)^5) \pmod {11} \\ & \equiv (1+9+4+3-10) \pmod {11} \\ & \equiv 7 \pmod {11} \end{aligned}$

Therefore, we have:

$\begin{aligned} a^4b^4 & \equiv 6^47^4 \pmod{11} \\ & \equiv 42^4 \pmod{11} \\ & \equiv (-2)^4 \pmod{11} \\ & \equiv \boxed{5} \pmod{11} \end{aligned}$