jmc

Let $\alpha$ be a root of $x^4+x^3+x^2+x+1=0,$ so $${\alpha }^4+{\alpha }^3+{\alpha }^2+\alpha +1=0.$$Then $(\alpha -1)({\alpha }^4+{\alpha }^3+{\alpha }^2+\alpha +1)=0,$ which simplifies to ${\alpha }^5=1.$ Then $$\begin{array}{rl}{\alpha }^{44}+{\alpha }^{33}+{\alpha }^{22}+{\alpha }^{11}+1& =({\alpha }^5{)}^8\cdot {\alpha }^4+({\alpha }^5{)}^6\cdot {\alpha }^3+({\alpha }^5{)}^4\cdot {\alpha }^2+({\alpha }^5{)}^2\cdot \alpha +1\\ & ={\alpha }^4+{\alpha }^3+{\alpha }^2+\alpha +1\\ & =0.\end{array}$$Since the first polynomial is 0 for every root $\alpha$ of the second polynomial, the first polynomial is divisible by the second polynomial, which means the remainder is $\boxed{0}.$