TITLE: ¿Is, for $1\le n<\omega ,\;(\omega+n)^{\omega}=\omega^{\omega}$?
QUESTION [6 upvotes]: As the title suggests, I am interested in calculating $(\omega+n)^{\omega}$.
My try goes like this: we know that:
$$(\omega+n)^{\omega}=\text{sup}\{(\omega+n)^k|\;k\in\omega\}$$
For instance, I would like to have some formula for $(\omega+n)^k$ which should be more manageable than its current form.
I proceed to prove, by induction on the natural numbers, that:
$$(\omega+n)^k=\omega^k+\dots+\omega^2n+\omega n+n$$
For if $k\in\omega$, and $k\not=0$, then:
$$(\omega+n)^{k+1}=(\omega^k+\dots+\omega^2n+\omega n+n)(\omega+n)=(\omega+n)^k=(\omega^k+\dots+\omega^2n+\omega n+n)\omega+(\omega^k+\dots+\omega^2n+\omega n+n)n$$
And it is not hard to prove, with another induction, that:
$$(\omega^k+\dots+\omega^2n+\omega n+n)n=\omega^kn+\dots+\omega^2n+\omega n+n$$
$$(\omega^k+\dots+\omega^2n+\omega n+n)\omega=\omega^{k+1}+\omega^{k-1}+\dots+\omega^2n+\omega n+n$$
And therefore;
$$(\omega^k+\dots+\omega^2n+\omega n+n)\omega+(\omega^k+\dots+\omega^2n+\omega n+n)n=\;=(\omega^{k+1}+\omega^{k-1}+\dots+\omega^2n+\omega n+n)+(\omega^kn+\dots+\omega^2n+\omega n+n)=\quad=\omega^{k+1}+(\omega^{k-1}+\dots+\omega^2n+\omega n+n+\omega^kn)+\dots+\omega^2n+\omega n+n=\qquad=\omega^{k+1}+\omega^kn+\dots+\omega^2n+\omega n+n$$
And therefore, $k+1$ also satisfies the condition, so the formula actually holds. However, what does:
$$\text{sup}\{\omega^k+\dots+\omega^2n+\omega n+n|\;k\in\omega\}$$
Actually equal to? Is it $\omega^{\omega}$? I am clueless at this point.
Thanks in advance for your time.

REPLY [3 votes]: Clearly $\omega^\omega\le(\omega+n)^\omega$. Also, $(\omega+n)^\omega\le (\omega^2)^\omega=\omega^{(2\cdot \omega)}=\omega^\omega$, and the equality follows.
If you do not feel comfortable with the move from $(\omega^2)^\omega$ to $\omega^{(2\cdot \omega)}$, simply note the left-hand side is $\omega\cdot\omega\cdot\omega\cdot\dots$, where there are $2\cdot\omega=\omega$ factors, all equal to $\omega$.