Let $K$ and $N$ be two languages. I want to prove that: $K^*N \cup N = K^*N$, where $*$ denotes the Kleene closure.
My idea is to say that I can write : $N={ε}N= K^0 N$.
Being $K^0$ the language $K$ raised to the power $0$ by definition all languages raised to the power equal a set with $ε$.
Replacing $N$ in $K^* N \cup N$ and using simplification of sets we have $K^*N \cup K^0N = K^*N$.
Is everything right here? Am I doing something that I can't? Thank you ((and I'm sorry I don't know how to put the $0$ up))
It is true that $K^0 = \{\epsilon\}$, but your sentence "using simplification of sets we have $K^*N \cup K^0N = K^*N$" hides the main argument.
Actually, you can simply observe that $N = \{\epsilon\}N$ and then, using distributivity, $$ K^*N \cup N = K^*N \cup \{\epsilon\}N = (K^* \cup \{\epsilon\})N $$ It remains to observe that $K^* \cup \{\epsilon\} = K^*$ since $\epsilon \in K^*$.