Finiteness of a $C^{*}$-algebra A unital $C^{*}$-algebra is said to be finite if $x^*x = 1$ implies $xx^* = 1$.
A unital $C^{*}$-subalgebra $B$ of a finite $C^{*}$-algebra $A$ is finite if $1_{B} = 1_{A}$.
Let $\lbrace A_n \rbrace$ be a sequence of finite unital $C^{*}$-algebras. Then both $\prod_{n=1}^{\infty}A_n$ and $\prod_{n=1}^{\infty}A_n / \oplus_{n=1}^{\infty}A_n$ are finte.
Could anybody help me to prove these two theorems?
 A: There is no magic happening here:
The unit of $\prod_nA_n$ is simply $(1_n)_{n\geq1}$ where $1_n$ denotes the unit of $A_n$. Now suppose that $(1_n)_{n\geq1}=x^*x$ for some $x\in\prod_nA_n$, say $x=(x_n)_{n\geq1}$, where $x_n\in A_n$ and $\sup_{n}\|x_n\|=\|x\|<\infty$. Then for each $n\geq1$ we have that $1_n=x_n^*x_n$ since the operations in $\prod_nA_n$ are pointwise. Since each $A_n$ is unital, we have that $x_nx_n^*=1_n$ as well, for all $n\geq1$. But this means that $1_{\prod_nA_n}=xx^*$. This shows that $\prod_nA_n$ is finite.
For the quotient, let $x\in\prod_nA_n$, $x=(x_n)$ so that $1+\sum_nA_n=(x+\sum_nA_n)^*\cdot(x+\sum_nA_n)$, thus $1-x^*x\in\sum_nA_n$. Equivalently, for each $\varepsilon>0$ there exists $n_0\geq1$ so that for all $n\geq n_0$ we have that $\|1_n-x_n^*x_n\|<\varepsilon$.

Lemma: If $p,q$ are projections in a $C^*$-algebra and $\|p-q\|<1$, then they are homotopic in the set of projections.

But homotopic projections are also Murray-von Neumann equivalent (see Rordam's book "An introduction to K-theory of $C^*$-algebras" for a proof of this and the above lemma). Now let $\varepsilon>0$. We have that there exists $n_0\geq1$ so that $\|1_n-x_n^*x_n\|<\min\{\varepsilon,1\}$ for all $n\geq n_0$. Therefore (by the lemma and our observation) we can find $y_n\in A_n$ so that $1_n=y_n^*y_n$ and $x_n^*x_n=y_ny_n^*$, for all $n\geq n_0$. But since $A_n$ is finite we have that $y_ny_n^*=1_n$, so $x_n^*x_n=1_n$ and again, since $A_n$ is finite we have that $x_nx_n^*=1_n$. This shows that $0=\|1_n-x_nx_n^*\|<\varepsilon$ for all $n\geq n_0$. In particular, $1-xx^*\in\sum_nA_n$, so $1+\sum_nA_n=(x+\sum_nA_n)\cdot(x+\sum_nA_n)^*$.
