Let $n,k\in\omega$. Then $\underbrace{(\omega+k)+(\omega+k)+\ldots+(\omega+k)}_{n\text{ times}}=\omega\cdot n+k$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Lemma: $k\in\omega\implies k+\omega=\omega$.
Proof: By definition, $k+\omega=\sup_{n\in\omega}(k+n)$ and $\omega=\sup_{n\in\omega}(n)$. It is clear that $\{k+n \mid n\in\omega\} \subseteq \{n \mid n\in\omega\}$ and that $\forall n\in\omega, \exists n'\in\omega:n\le k+n'$. The result is then followed.
We proceed to prove our main theorem by induction on $n$.
The statement is trivially true for $n=1$.
Assume that $\underbrace{(\omega+k)+(\omega+k)+\ldots+(\omega+k)}_{n\text{ times}}=\omega\cdot n+k$.
Then $\underbrace{(\omega+k)+(\omega+k)+\ldots+(\omega+k)}_{n+1\text{ times}}$
$=\underbrace{(\omega+k)+(\omega+k)+\ldots+(\omega+k)}_{n\text{ times}}+(\omega+k)=(\omega\cdot n+k)+(\omega+k)$
$=\omega\cdot n+(k+(\omega+k))=\omega\cdot n+((k+\omega)+k)\overset{\mathrm{Lemma}}{=}\omega\cdot n+(\omega+k)=(\omega\cdot n+\omega)+k=\omega\cdot (n+1)+k$.
This completes the proof.