Squares of C*-algebras I'm reading a paper where it is claimed that every C*-algebra $A$ satisfies $A^2 = A$, "for example, using Cohen's 1959 factorization theorem". However, I don't see how to apply Cohen's factorization theorem to obtain this. 
Is there an easier proof that doesn't use Cohen's factorization theorem?
 A: It is a direct consequence of the fact that any C$^*$-algebra has an approximate identity. Then Cohen's Factorization Theorem tells you directly that $A\subset A^2$. 
It can also be obtained directly from the following result: 

For any $a\in A$, there exists $b\in A$ such that $a=b(a^*a)^{1/4}\in A^2$.

Proof. (from Davidson's book) Let $b_n=a(a^*a+\frac1n\,I)^{-1/2}(a^*a)^{1/4}$. This makes sense on the unitization of $A$, but lies in $A$. Then, with $f_n(t)=t^{3/4}(t+\frac1n)^{-1/2}$, we have $f_n\to t^{1/4}$ uniformly on $[0,\|a^*a\|]$ and
$$
\|b_n-b_m\|^2=\|a\left[(a^*a+\frac1n\,I)^{-1/2}-(a^*a+\frac1m\,I)^{-1/2}\right]\,(a^*a)^{1/4}\|^2\\
=\|(a^*a)^{1/4}\left[(a^*a+\frac1n\,I)^{-1/2}-(a^*a+\frac1m\,I)^{-1/2}\right]a^*a\left[(a^*a+\frac1n\,I)^{-1/2}-(a^*a+\frac1m\,I)^{-1/2}\right]\,(a^*a)^{1/4}\|\\
=\|(a^*a)^{1/2}\left[(a^*a+\frac1n\,I)^{-1/2}-(a^*a+\frac1m\,I)^{-1/2}\right]\,(a^*a)^{1/4}\|^2\\
(\text{for two selfadjoints }x,y,\ \|xy\|=\|yx\|)\\
=\|\left[(a^*a+\frac1n\,I)^{-1/2}-(a^*a+\frac1m\,I)^{-1/2}\right]\,(a^*a)^{3/4}\|^2\\
=\|f_n(a^*a)-f_m(a^*a)\|^2\leq\|f_n-f_m\|_{[0,\|a^*a\|]}^2.
$$
Thus the sequence $\{b_n\}$ is Cauchy. Let $b=\lim_nb_n$. Then
$$
b(a^*a)^{1/4}=\lim_nb_n(a^*a)^{1/4}=\lim_n\, a(a^*a+\frac1n\,I)^{-1/2}(a^*a)^{1/2}=a
$$
