If $\lambda$ is a limit ordinal , $0\le n < \omega$, $1<m<\omega$
Prove that $(\lambda+n)^{m}<\lambda^{m}*2$ and deduce that $(\lambda+n)^{\omega}=\lambda^{\omega}$.
My attempt at the first proof is as follows;
By induction on $n$, when $n=0$ we get that $(\lambda+0)^{m}=\{\sup_{\epsilon<\lambda}\epsilon^m\}+0=\lambda^{m}+0<\lambda^{m}*2$ so the case when $n=0$ holds.
Now let $n=\delta+1$ then $(\lambda+(\delta+1))^{m}=\lambda^{m}+(\delta+1)^{m} =\lambda^{m}+n^{m}<\lambda^{m}*2$, since $\lambda>n$ as $\lambda$ is a limit ordinal.
Now when $n$ is a limit ordinal we have that $(\lambda+n)^{m}=\lambda^{m}+\{\sup_{\epsilon<n}\epsilon^{m}\}=\lambda^{m}+n^{m}<\lambda^{m}*2$. End of proof.
Now i'm unsure where to approach from to deduce the other required result, would I just again proceed with induction on $n$ the trivial case when $n=0$ seems to give the required result, however in my proof above i'm unsure if my successor step is correct or my limit case in fact so if someone could explain the problem and where i've gone wrong (if i have) I would be very thankful. I'm mostly looking for a solution that just uses transfinite induction.
First $$ \lambda^{m}<(\lambda+n)^m<\lambda^{m}\cdot2<\lambda^{m}\cdot\lambda^{m}=\lambda^{2m} $$ So $$ \lambda^{\omega}<(\lambda+n)^{\omega}<\lambda^{2\omega}=\lambda^{\omega} $$ Also your proof using transfinite induction is not right. The right proof should be on $m$ and is as follows. For $m=1$, it is obvious that $$ \lambda+n<\lambda+\lambda=\lambda\cdot2 $$ For $m=\delta +1$ \begin{align} (\lambda+n)^{\delta+1}&=(\lambda+n)^{\delta}(\lambda+n) \\ &<(\lambda^{\delta}\cdot2)(\lambda+n) \\ &=(\lambda^{\delta}\cdot2)\lambda+(\lambda^{\delta}\cdot2)n \\ &=\lambda^{\delta}\cdot2\lambda+\lambda^{\delta}\cdot2n \\ &<\lambda^{\delta}\cdot2\lambda+\lambda^{\delta}\cdot2\lambda \\ &=\lambda^{\delta}\cdot\lambda+\lambda^{\delta}\cdot\lambda \\ &=\lambda^{\delta+1}\cdot2 \end{align} If $m$ is limit ordinal $$ (\lambda+n)^m=\bigcup_{\alpha<m}(\lambda+n)^{\alpha}<\bigcup_{\alpha<m}\lambda^{\alpha}\cdot2=\lambda^{m}\cdot2 $$