Lie group structure of the spin group let $Cl_n:=T(\mathbb{R}^n)/I$ be the clifford algebra of $\mathbb{R}^n$ with the standard inner product. (Here $T(\mathbb{R}^n)$ denotes the tensor algebra of $\mathbb{R}^n$ and $I$ is the ideal genereted by all elements of the form $x\otimes x+<x,x>1$, $x\in\mathbb{R}^n$.)
Then $Spin_n:=\{x_1\cdot\ldots\cdot x_{2k}| x_j\in\mathbb{R}^n, |x_j|=1,k\in\mathbb{N}\}\subset Cl_n^*$ is called the spin group.
I think I got the basic idea of how to show that $Spin_n$ is a lie group, but I don't understand some of the details. These I want to ask here.
First of all, I came across the following statement:

Lemma 1: Let $A$ be a finite dimensional associative algebra over $\mathbb{R}$. Then the multiplicative group of invertible elements $A^*$ is open in $A$ and a lie group.

If you combine that with

Lemma 2: $Spin_n\subset Cl_n^*$ is closed

you obtain that $Spin_n$ is a lie group (because it's a lie subgroub of the lie group $Cl_n^*$).

Question: How do I prove Lemma 1 and Lemma 2?

My thoughts on the proof of Lemma 1:
First of all, $A$ is a finite dimensional real vector space, thus a smooth manifold. Define a map $$\lambda\colon A\rightarrow End(A), a\mapsto(v\mapsto a\cdot v).$$ $\lambda$ is linear, so it's smooth. The subset $GL(A)\subset End(A)$ is open. So $A^*$ would be open if 
$$\lambda^{-1}(GL(V))=A^*.$$
I'm not sure if that holds ($\subseteq$ is the problem).
If $A^*\subset A$ is open, it is a smooth manifold, but why is the the map $A^*\rightarrow A^*, a\mapsto a^{-1}$ smooth?
 A: On account of Lemma 1: 
Equip $A$ with a submultiplicative norm $\| \cdot \|$ making it into unital Banach algebra. In this case you can use the usual infinite sum to find the inverse of element $1 + a \in A$ for $\| a \| < 1$. Using this trick, you can show that arount each $a_{0} \in A^{\ast}$, there is an open ball $B_{r(a_{0})}(a_{0}) \subseteq A^{\ast}$ where all elements are invertible. Hence $A^{\ast}$ is open in $A$.
Multiplication in $A$ is bilinear, hence smooth, and it thus restricts to the smooth multiplication in $A^{\ast}$. It remains to prove that the inverse is a smooth map. This is easy to prove in the unit ball $B_{1}(1)$ around the unit $1 \in A$, where you can invert using the power series - and absolutely convergent power series is smooth as all its partial sums are polynomials in $a \in A$. Hence the inverse is smooth on $B_{1}(1)$. For any $a_{0} \in A^{\ast}$, you can again find open ball $B_{r'(a_{0})}(a_{0}) \subseteq A^{\ast}$ which is mapped by left translation (smooth map) into $B_{1}(1)$. You can then invert any element in this ball by left-translating, inverting in $B_{1}(1)$, right translating. This proves that point $a_{0} \in A^{\ast}$ has an open neighbourhood where the inversion is smooth.
On account of Lemma 2: 
Maybe this can be proved directly, but to my understanding, usual proof is to first consider closed subgroup of elements fixing the subspace of $Cl^{1}_{n}$ under the twisted adjoint action of $Cl_{n}^{\ast}$ on $Cl_{n}$, which is usually called the Clifford group $\Gamma_{n}$. Its intersection with even part of $Cl_{n}$ is called a special Clifford group $\Gamma_{n}^{+}$. Groups $Pin_{n}$ and $Spin_{n}$ are then obtained as a preimage of $\{1\} \in \mathbb{R}^{\ast}$ of a certain smooth map $N: \Gamma_{n} \rightarrow \mathbb{R}^{\ast}$ of $N^{+}: \Gamma_{n}^{+} \rightarrow \mathbb{R}^{\ast}$, respectively. See for example http://www.cis.upenn.edu/~cis610/clifford.pdf for a definition of $N$, twisted adjoint action, et cetera. But preimage of a point is certainly closed, hence $Pin_{n}$ and $Spin_{n}$ are closed subgroups of $\Gamma_{n}$ and $\Gamma_{n}^{+}$, respectively.
