Could you proof it to me that A(A+B) = A?
AA + BA [AA = A]
A + AB
Then what?
I presume you are looking for a way to prove the identity using a calculus. So far you have used distributivity an idempotency.
Recall that A = A1
so you get A1+AB
and you can use distributivity again, this time in the other direction. Then two obvious steps.
Here is a proof: $$ A \stackrel{(1)}= A \cdot 1 \stackrel{(2)}= A \cdot (1+B) \stackrel{(3)}= A \cdot 1 + A \cdot B \stackrel{(4)}= A + A \cdot B \stackrel{(5)}= A \cdot A + A \cdot B \stackrel{(6)}= A \cdot (A+B). $$
Axioms used:
(1),(4) multiplicative identity
(2) absorption
(3),(6) distributivity
(5) idempotence
I always liked the $\min, \max$ definitions of $\cdot$ and $+$, since some courses in boolean algebra just give those laws and ask you to accept them. [Boolean Algebra: Basic Operations]
\begin{align*} x \land y &= x \cdot y = \min(x,y)\\ x \lor y &= x + y = \max(x,y) \end{align*} where $0 \leq x,y \leq 1$.
So $A(A+B)$ becomes: $$ \min(A, \max(A,B)) $$
\begin{align*} \min(A, \color{red}{\max(A,B)}) &= \min(A, \color{red}{B})\\ &= \min(A, B) = A\\ \end{align*}