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In answer to the second question, yes this is true. Say $x^k=0$. Let $x=v|x|$ be the polar decomposition of $x$ in $A^{**}$. Let $\tilde x=|x|^{\frac 1 2}v|x|^{\frac 1 2}$ (the Aluthge transform of $x$). Then $$ \tilde x^{k-1}(\tilde x^{k-1})^*=|x|^{\frac 1 2}x^{k-1} v^*(x^{k-2})^*|x|^{\frac 1 2}=0, $$ where we have used that $|x|^{\frac 1 2}x^{k-1}=0$ (since $|x|^{\frac 1 2}\in C^*(x^*x)$ and $(x^*x)x^{k-1}=0$). Hence $\tilde x^{k-1}=0$.