6.6.48 Combinatorics

Condition: determine \ (a^{n}: a^{1} = A \ Weedge a^{K+1} = A^{K} \ CDOT A \). Prove: a) \ (a^{m+n} = a^{m} \ cdot a^{n} \), b) \ ((a \ cdot b)^{m} = a^\ cdot b^{m} \).