please help :DDD

If we add all the exponents up, we get 3^(1+2+3+4... 43) = 3^(946). 

3, 9, 27, 81, 243, 729...

3, 9, 5, 4, 1, 3....

There is a cycle of 3, 9, 5, 4, 1 in mod 11. 

946 = 1 mod 11, and the first number in the cycle of 3, 9, 5, 4, 1 is 3. 

=^._.^=

Find the remainder when $3*3^2*3^3*\ldots*3^{43}$ is divided by $11$.

$\begin{array}{|rcll|} \hline 3*3^2*3^3*\ldots*3^{43} &=& 3^{1+2+3+\ldots 42+43} \\ &=& \left.3\right.^{\left(\frac{1+43}{2}*43\right)} \\ &=& \left.3\right.^{\left(\frac{44}{2}*43\right)} \\ &=& 3^{22*43} \\ &=& 3^{946} \\ \hline \end{array}$

$\begin{array}{lrcll} \text{Euler:}\\ \boxed{3^{\phi(11)}\equiv 1 \pmod{11} \quad | \quad \phi(11)=10,~ \phi(p) = p-1\\3^{10}=1 \pmod{11} } \\ \end{array}$

$\begin{array}{|rcll|} \hline 3^{946} \pmod{11} &\equiv& 3^{10*94+6}\pmod{11} \\ &\equiv& \left(3^{10} \right)^{94}3^6\pmod{11} \quad | \quad 3^{10} \equiv 1\pmod{11} \\ &\equiv& 1^{94}3^6\pmod{11} \\ &\equiv& 3^6\pmod{11} \\ &\equiv& 729\pmod{11} \\ \mathbf{3^{946} \pmod{11}} &\equiv& \mathbf{ {\color{red}3}\pmod{11}} \\ \hline \end{array}$