There is a well ordering of the class of all finite sequences of ordinals

For $f,g\in\operatorname{Ord}^{<\omega}$ define $n_{f,g}$ to be the value $\min\{n\mid f(n)\neq g(n)\}$ (if $f\neq g$). Now you can define $f<g$ to mean this: \begin{align*} &\max\operatorname{ran}(f) < \max\operatorname{ran}(g)\\ \lor&(\max\operatorname{ran}(f) = \max\operatorname{ran}(g) \land \operatorname{dom}(f) < \operatorname{dom}(g))\\ \lor&(\max\operatorname{ran}(f) = \max\operatorname{ran}(g) \land \operatorname{dom}(f) = \operatorname{dom}(g) \land f(n_{f,g}) < g(n_{f,g})) \end{align*} This is the well-ordering you're looking for. Now prove that it does what it's supposed to do!

Hint: Let $A_g$ be the initial segment $\{ f \in \operatorname{Ord}^{<\omega} \mid f < g \}$. It is easy to see that for every $\alpha$ the set $A_{\{(0,\alpha)\}}$ is equal to $\alpha^{<\omega}$; specifically - replacing $\alpha$ with $\omega_\alpha$ -, each $\omega_\alpha^{<\omega}$ is an initial segment in $<$.

Because of $|\omega_\alpha|\leq|\omega_\alpha^{<\omega}|$, the order type of $\omega_\alpha^{<\omega}$ can't be smaller than $\omega_\alpha$. Now suppose there's a smallest $\alpha$ such that the order type of $\omega_\alpha^{<\omega}$ is strictly greater than $\omega_\alpha$ and show that this leads to a contradiction.