I started to learn von Neumann algebras and I wonder about the proof of the following. Can anyone outline it?
If $\tau$ is a trace on a von Neumann algebra $M$, in other words $M$ is $II_1$-factor, then for every $a\in M$ operator we have
$$\|a\|=\lim \tau[(a^*a)^n]^{2/n}.$$
Note first that a von Neumann algebra with a trace is not necessarily a II$_1$-factor. An infinite-dimensional vN algebra with a unique faithful trace is indeed a II$_1$-factor.
Regarding your question, it looks to me that your exponents are off. But we'll see about that. Suppose that $b\in M^+$, with $\|b\|=1$. Fix $\varepsilon>0$. Then there exists a projection $p\in M$ such that $b\geq(1-\varepsilon)\,p$. If $\tau$ is faithful, then $\tau(p)>0$. Thus $$ \tau(b^n)\geq(1-\varepsilon)^n\,\tau(p), $$ and then $$ \tau(b^n)^{1/n}\geq(1-\varepsilon)\,\tau(p)^{1/n}, $$ implying $$ \liminf_n\tau(b^n)^{1/n}\geq(1-\varepsilon)\,\liminf_n\tau(p)^{1/n}=1-\varepsilon. $$ As we can do this for any $\varepsilon>0$, we get $\liminf_n\tau(b^n)^{1/n}\geq1$. On the other hand, $$ \tau(b^n)^{1/n}\leq\tau(\|b\|^n)^{1/n}=1. $$ So $1=\lim_n\tau(b^n)^{1/n}$. For other positive $b$, we apply the above to $b/\|b\|$, and so $\|b\|=\lim_n\tau(b^n)^{1/n}$. Finally, for arbitrary $a\in M$, $$ \|a\|^2=\|a^*a\|=\lim_n\tau((a^*a)^n)^{1/n}. $$
Edit: about the uniqueness of the trace. If $z\in Z(M)$ is any positive element in the centre of $M$, then $\tau(z\cdot)$ is a trace. An invertible positive element in the centre will lead to a faithful trace. So if $M$ has a unique faithful trace, let $z\in Z(M)$ be invertible and positive. Then $x\mapsto \tau(zx)$ is a faithful trace. By the uniqueness, $\tau(zx)=\tau(x)$ for all $x$, or $\tau((1-z)x)=0$ for all $x$. Taking $x=1-z$, the faithfulness of the trace implies $z=1$. So the centre is trivial and $M$ is a factor.
Now $M$ is an infinite-dimensional factor with a faithful trace, so it is a II$_1$-factor.