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 $A^{**}$ (the bidual of $A$). Let $a=v|x|^{\frac 1 2}$ and $b=|x|^{\frac 1 2}$. Then clearly $x=ab$. Both $a$ and $b$ belong to $A$. In $b$'s case, by functional calculus. It is a well-known property of polar decompositions that $v|x|^{\frac 1 2}$ is also in $A$. To see this, write $p_n(|x|)\to |x|^{\frac 1 2}$, where each $p_n$ is a polynomial such that $p_n(0)=0$. Then $vp_n(|x|)\to v|x|^{\frac 1 2}$ in norm and $vp_n(|x|)\in A$ for all $n$ because we can factor out $|x|$ from $p_n(|x|)$.
Now consider $ba=|x|^{\frac 1 2}v|x|^{\frac 1 2}$ (the Aluthge transform of $x$). Then $$ (ba)^{k-1}(ba)^*= (|x|^{\frac 1 2}v|x|^{\frac 1 2}\cdots |x|^{\frac 1 2}v|x|^{\frac 1 2})\cdot |x|^{\frac 1 2}v^*|x|^{\frac 1 2}= |x|^{\frac 1 2}x^{k-1} v^*|x|^{\frac 1 2}=0, $$ where we have used that $|x|^{\frac 1 2}x^{k-1}=0$ (since $|x|^{\frac 1 2}\in C^*(x^*x)$ and $(x^*x)x^{k-1}=0$). It follows that $(ba)^{k-1}((ba)^{k-1})^*=0$ which implies that $(ba)^{k-1}=0$.