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Binomial theorem expansion so we have $\sum _{i=0}^4\binom{4}{i}\cdot \:1^{\left(4-i\right)}i^i=1+4i-6-4i+1=\boxed{-4}~~\blacksquare$

Note that i is additionally the index of the summation.

Compute  $\large{\left(1 + i\right)^4}$

$\begin{array}{|rcll|} \hline && \mathbf{\left(1 + i\right)^4} \\ &=& \Big(~\left(1 + i\right)^2~\Big)^2 \quad &| \quad \left(1 + i\right)^2 = 1+2i+i^2 \\ &=& \Big(~1+2i+i^2~\Big)^2 \quad &| \quad i^2=-1 \\ &=& \left(~1+2i-1~\right)^2 \\ &=& \left( 2i \right)^2 \\ &=&4i^2\quad &| \quad i^2=-1 \\ &=&4(-1) \\ &=& \mathbf{-4} \\ \hline \end{array}$

$\text{Just as an alternative way to calculate this}\\ z = 1+i = \sqrt{2} e^{i \pi/4}\\ z^4 = (\sqrt{2})^4 e^{4(i\pi/4)} = 4e^{i\pi} = -4$