6.6.48 Combinatorics

Condition: Let us define \( a^{n}: a^{1}=a \wedge a^{k+1}=a^{k} \cdot a \). Prove: a) \( a^{m+n}=a^{m} \cdot a^{n} \), b) \( (a \cdot b)^{m}=a^{m} \cdot b^{m} \).