Prove that $\alpha\ge2\Rightarrow\alpha^{\beta}\ge\alpha\beta$.
Let's use transfinite induction.
- $P(0)$
$\alpha^{0}\ge\alpha\cdot0\iff1\ge0$.
- $P(\beta)\Rightarrow P(\beta+1)$
Let's multiply $\alpha^{\beta}\ge\alpha\beta$ by $\alpha$: $\alpha^{\beta+1}\ge\alpha\beta\alpha\ge\alpha(\beta+1)$
$(\beta\alpha\ge\beta+1$ with $\alpha\ge2$
).
Another way for $\beta\alpha\ge\beta+1$: $$\beta+\beta\alpha\ge\beta+\beta+1$$ $$\beta(1+\alpha)\ge\beta(2+1)$$ $$1+\alpha\ge 3$$ is true when $\alpha\ge2$, naturally.
- $P(\beta)\Rightarrow P(\gamma)$
($\gamma$ is a limit ordinal such that $\gamma>\beta$).
$\gamma>\beta \Rightarrow \gamma=\beta+\sigma.$ Let's multiply $\alpha^{\beta}\ge\alpha\beta$ by $\alpha^{\sigma}$: $\alpha^{\gamma}=\alpha^{\beta+\sigma}\ge\alpha\beta\alpha^{\sigma}\ge\alpha(\beta+\sigma)=\alpha\gamma$. We need to verify $\beta\alpha^{\sigma}\ge\beta+\sigma$:
$$\beta\alpha^{\sigma}\ge\beta+\sigma$$ $$\beta+\beta\alpha^{\sigma}\ge\beta+\beta+\sigma$$ $$\beta(1+\alpha^{\sigma})\ge\beta(2+\sigma)$$ $$1+\alpha^{\sigma}\ge2+\sigma$$ $$\alpha^{\sigma}\ge1+\sigma$$ From $\beta<\gamma$, $\gamma=\beta+\sigma$, and the fact that gamma is a limit ordinals while beta is not, I think, follows that $\sigma$ is also a limit ordinal. So, also considering the smallest possible alpha (the inequation would then naturally hold for any bigger ones), $$2^{\sigma}\ge\sigma,$$ which is true ($2^{\omega}=\omega$, for example).
I'd like to know whether my proof is correct or not, and if it is, whether it's rigourous enough. It would also be great to see alternative ways of proving this.
Thank you in advance.
The drawing is definitely not a proof of $\beta \alpha \geq \beta+1$. In particular, the graph you drew is of a function of natural numbers and says nothing about a function of ordinals.
Also, $\beta + \beta +1 \neq \beta(2+1)$, so that part is also wrong.