I'm trying to learn about transfinite ordinals and got stuck here.
Once you've added $1$ infinitely, you can add another $1$ and get a larger result. If you take the successor of $$1+1+1+1+\dots = \omega,$$ the result is $(\omega+1)$, larger than $\omega$.
Once you've added a limit ordinal to itself infinite times, you can add it again and get a larger result. If you add $\omega$ to $\omega$ already added to itself infinitely $$\omega+\omega+\omega+\omega+\dots= \omega^2,$$ you get $\omega^2+\omega$, larger than $\omega^2$.
Once you've multiplied a limit ordinal by itself infinite times, the pattern holds. Multiply $\omega$ with itself infinitely $$\omega\cdot\omega\cdot\omega\cdot\omega\cdot\dots = \omega^\omega$$ and then multiply $\omega$ for $\omega^\omega\cdot\omega > \omega^\omega$.
But when exponentiating infinitely, it breaks. If you exponentiate $\omega$ to $\omega^{\omega^{\omega^{\omega^{\dots}}}}$, nothing changes and you get back $\omega^{\omega^{\omega^{\omega^{\dots}}}}$.
Why is that? Why isn't $\omega\uparrow\uparrow(\omega+1$ larger? How does one prove that $\omega\uparrow\uparrow(\omega+1)$ contains/is a subset of $\omega\uparrow\uparrow\omega $?
Look at it this way:
$1+\omega=\omega$ and $\omega$ is the least ordinal $\alpha$ with $1+\alpha=\alpha$.
$\omega+\omega^2=\omega^2$ and $\omega^2$ is the least $\alpha$ with $\omega+\alpha=\alpha$.
$\omega\cdot \omega^\omega=\omega^\omega$ and $\omega^\omega$ is the least $\alpha$ with $\omega\cdot\alpha=\alpha$.
$\omega^{\varepsilon_0}=\varepsilon_0$ and $\varepsilon_0$ is the least $\alpha$ with $\omega^\alpha=\alpha$.
But $\omega+1>\omega$, $\omega^2+\omega>\omega^2$, $\omega^\omega\cdot\omega>\omega^\omega$ as you mention and similarly $\varepsilon_0^\omega>\varepsilon_0$. Your concern stems from treating the order of operations differently in the case of exponention towers compared to addition, multiplication etc.