I am following Jech's construction, by definition $\alpha+0 = \alpha, \alpha+(\beta+1)=(\alpha+\beta)+1$, and for limit $\beta$ we define $\alpha+\beta = \cup\{\alpha+\xi: \xi<\beta\}$.
Jech's proof of associativity just says "By induction on $\gamma$." so I am trying to go through it. The statement I am inducting on is $\forall \alpha,\beta \in Ord~ (\alpha+(\beta+\gamma) = (\alpha+\beta)+\gamma)$.
I have done $\gamma = 0$ and $\gamma=c+1$ and have left to show it holds for $\gamma$ a limit. In this case I would like to say $$(\alpha+\beta)+\gamma = \cup\{(\alpha+\beta)+\xi:\xi < \gamma\}$$ $$ = \cup\{\alpha+(\beta+\xi):\xi < \gamma\} $$ $$ = \alpha + \cup\{\beta+\xi:\xi<\gamma\} $$ $$ = \alpha + (\beta+\gamma). $$
However I have been stuck trying to show $ \cup\{\alpha+(\beta+\xi):\xi < \gamma\} = \alpha + \cup\{\beta+\xi:\xi<\gamma\}. $
How can I proceed?
For establishing your last statement, it suffices to prove two things (neither is hard):
For, the right-hand side of this equality is precisely the definition of $\alpha+(\beta+\gamma)$ when $\beta+\gamma$ is a limit ordinal.
Alternatively, you can take the general route and prove that the composition of normal sequences is normal, and directly conclude that:
$$\sup_{\xi <\gamma}(\alpha +(\beta+\xi)) = \alpha+(\beta+\gamma)$$
because the sequences $\delta_\zeta = \alpha+\zeta$ and $\epsilon_\xi = \beta+\xi$ are both normal, and thus so is the sequence $\eta_\xi = \delta_{\epsilon_\xi} = \alpha + (\beta + \xi)$. This approach will save you time when dealing with the multiplication and exponentiation cases.