Prove that $(\omega+n)^\omega=\omega^\omega$ where $n$ is natural number.

Question

I want to prove that $(\omega+n)^\omega=\omega^\omega$ where $n$ is natural number.

This is probably trivial, but I am not sure if I know how to properly justify it. I am using following definition of ordinal exponential: $$ \alpha^\beta=tp(F(\alpha,\beta),\trianglelefteq) $$ where $F(A,B)$ is set of functions from $B$ to $A$ with finite support. and $f\trianglelefteq g \iff f=g \land f(a)<f(a)$ where $a=\max\{b\leq\beta:f(b)\neq g(b)\}$.

So we have: $$ (\omega+n)^\omega=(tp(\{0\}\times\omega\cup\{1\}\times n, \le))^\omega=tp(F(\{0\}\times\omega\cup\{1\}\times n,\omega)) $$

So now we want to know how many function with finite support from $\omega$ to $F(\{0\}\times\omega\cup\{1\}\times n)$ there are. We can ignore order for such considerations.

Therefore we only need to know how many functions with finite support are from set $B$ to set $A$ where $B$ has $|\omega|$ elements and $A$ has $|n|+|\omega|=|\omega|$ elements.

Is it correct reasoning?

Answer 1

Use the fact that $$\alpha^{\lambda}=\sup\limits_{\beta<\lambda} \alpha^{\beta}$$ as from for example the recursive definition of ordinal exponentiation. Now $$\omega <\omega +n<\omega^2$$ thus for all $k$, $$\omega^k <(\omega +n)^k<\omega^{2k}$$ taking the sup we get $$\omega^{\omega} \leq(\omega +n)^{\omega}\leq\omega^{\omega}$$