Godel's Constructible Universe is defined by transfinite recursion as follows:
$$ L_{0}:=\varnothing .$$ $${\displaystyle L_{\alpha +1}:=\operatorname {Def} (L_{\alpha }).} $$ If ${\displaystyle \lambda } $ is a limit ordinal, then: $$L_{\lambda }:=\bigcup _{\alpha <\lambda }L_{\alpha }$$ and: $${\displaystyle L:=\bigcup _{\alpha \in \mathbf {Ord} }}L_{\alpha}$$ Where
$\operatorname {Def} (X):={\Bigl \{}\{y\mid y\in X{\text{ and }}(X,\in )\models \Phi (y,z_{1},\ldots ,z_{n})\}~{\Big |}~\Phi {\text{ is a first-order formula and }}z_{1},\ldots ,z_{n}\in X{\Bigr \}}.$
I'm reading through Jech's 'Axiom of Choice' on the Constructible Universe and am trying to understand why $$(V= L)^L$$ The proof is to show that '$x$ is constructible' is an absolute formula, and then to state that $$(V=L)^L \iff (V=L)^V$$ By which the result follows by absoluteness. However this is the problem I'm having: Surely, (a) It is trivially true that $L$ believes that all sets are constructible, (how on earth could it not?) and (b) surely it is trivially false that $(V=L)^V$, after all, $V$ is constructed without the stipulation that sets must be definable, whereas $L$ is. Am I missing something here?
No. This is not trivial at all.
The point here is that $x\in L$ is a $\Delta_1$ formula, so it is indeed absolute and $L^L=L^V=L$.
But this is not necessarily true for other classes, even if they seem "very definable". For example, the class $\rm HOD$, which is the class sets which are hereditarily definable from the ordinals, is also an inner model of $\sf ZFC$, but $\rm HOD^{HOD}$ is not necessarily $\rm HOD^V$. In fact, there can be an inner model $M$ such that $\mathrm{HOD}\subsetneq\mathrm{HOD}^M$.
So the fact that the formula for $L$ is "simple" is a very important fact.
Secondly, there is the point that $V$ and $L$ are constructed in very different ways. And even if you look at the von Neumann hierarchy inside $L$, it will still be very different from the constructible hierarchy itself (although the two will often coincide).