Problem
Given a C*-algebra with unit $1\in\mathcal{A}$.
Define positive elements as: $$A\geq0\iff\sigma(A)\geq0\quad(A=A^*)$$
Positive elements can be characterized by: $$A\geq0\iff A=B^*B$$
Attempts
One direction easily follows from the continuous calculus: $$A\geq0\implies A=\sqrt{A}\sqrt{A}$$
For operator algebras the numerical range becomes accessible: $$A=B^*B\implies\mathcal{W}(A)\geq0\implies\sigma(A)\geq0$$
For general C*-algebras one has a faithful representation : $$\pi:\mathcal{A}\to\mathcal{B}(\mathcal{H}):\quad\ker\pi=(0)$$
But how to show the above without exploiting advanced tools like Gelfand Naimark?
Ok, I guess I have it...
First, let's do some recapitulation...
Relations
By the C*-property one has: $$A=A^*:\quad\|A\|=r(A)$$ Remember also that in a ring: $$\sigma(AB)\cup\{0\}=\sigma(BA)\cup\{0\}$$
Decompositions
Every element has a decomposition into selfadjoints: $$A=\Re A+\imath\Im A\quad(\Re A=\Re A^*,\,\Im A=\Im A^*)$$ and every selfadjoint has a decomposition into positive ones: $$A=A^*:\quad A=A_+-A_-\quad(A_\pm\geq0)$$ Especially they cancel each other: $A_+A_-=A_-A_+=0$
Now, let's tackle the problem...
Combos
Observe that the order is definit: $$0\leq A\leq0\implies\|A\|=r(A)=0\implies A=0$$ Also, one has the noncommutative complex analog: $$0\leq\Re A^2+\Im A^2=A^*A+AA^*$$ Thus, combos cannot be negative: $A^*A\leq0\implies A^*A=0$
Exploiting this one deduces: $$\left(A\sqrt{(A^*A)_-}\right)^*\left(A\sqrt{(A^*A)_-}\right)=(A^*A)_-(A^*A)=-(A^*A)_-^2\leq0\\\implies-(A^*A)_-^2=\left(A\sqrt{(A^*A)_-}\right)^*\left(A\sqrt{(A^*A)_-}\right)=0\implies A^*A=(A^*A)_+\geq0$$
Concluding, combos are positive: $A^*A,AA^*\geq0$