[proofplan]
We work inside $L$ and prove the upper bound $|\mathcal{P}(\kappa)\cap L|^L \leq \kappa^+$ for an arbitrary infinite $L$-cardinal $\kappa$. Given $X \subseteq \kappa$ with $X \in L$, we place $X$ in a sufficiently large level $L_\theta$, take an elementary substructure of size $\kappa$ containing $X$ and all ordinals below $\kappa+1$, and collapse it. The Condensation Lemma identifies the collapse with some $L_\beta$ where $\beta < \kappa^+$, so every constructible subset of $\kappa$ already appears before level $L_{\kappa^+}$. Since $L_{\kappa^+}$ has cardinality $\kappa^+$ in $L$, this gives the upper bound, while [Cantor's theorem](/theorems/621) gives the reverse inequality.
[/proofplan]
[step:Work in $L$ and fix the cardinal whose power set is being counted]
Work in the universe $L$. Let $\kappa$ be an infinite cardinal of $L$, and let $\kappa^+$ denote the successor cardinal of $\kappa$ computed in $L$. Define
\begin{align*}
\mathcal{P}_L(\kappa) := \{X \in L : X \subseteq \kappa\}.
\end{align*}
It is enough to prove that $|\mathcal{P}_L(\kappa)|^L = \kappa^+$. Since $L$ satisfies the axioms of choice, cardinal comparison in $L$ is linear, so it suffices to prove both inequalities
\begin{align*}
\kappa^+ \leq |\mathcal{P}_L(\kappa)|^L
\qquad\text{and}\qquad
|\mathcal{P}_L(\kappa)|^L \leq \kappa^+.
\end{align*}
[/step]
[step:Use Cantor's theorem to obtain the lower bound]
By Cantor's theorem in $L$ (citing a result not yet in the wiki: Cantor's theorem), there is no surjection in $L$ from $\kappa$ onto $\mathcal{P}_L(\kappa)$. Hence
\begin{align*}
\kappa < |\mathcal{P}_L(\kappa)|^L.
\end{align*}
Because $\kappa^+$ is the least $L$-cardinal strictly larger than $\kappa$, it follows that
\begin{align*}
\kappa^+ \leq |\mathcal{P}_L(\kappa)|^L.
\end{align*}
[/step]
[step:Show every constructible subset of $\kappa$ appears before $L_{\kappa^+}$]
We prove that
\begin{align*}
\mathcal{P}_L(\kappa) \subseteq L_{\kappa^+}.
\end{align*}
Let $X \in \mathcal{P}_L(\kappa)$. Since $X \in L$, there is an ordinal $\theta$ such that $X \in L_\theta$. Increase $\theta$ if necessary so that $\theta > \kappa^+$ and $L_\theta$ satisfies a sufficiently large finite fragment of ZFC for the Downward Löwenheim-Skolem argument and the Condensation Lemma.
Apply the [Downward Löwenheim-Skolem theorem](/theorems/4272) inside $L$ (citing a result not yet in the wiki: [Downward Löwenheim-Skolem theorem](/theorems/4299)) to the structure $(L_\theta,\in)$ with the parameter set
\begin{align*}
A := \{X\} \cup (\kappa+1).
\end{align*}
Since $\kappa$ is infinite, $\kappa+1$ has $L$-cardinality $\kappa$, and adjoining the single element $X$ does not increase that cardinality. Hence $A$ has $L$-cardinality $\kappa$. Thus there is an elementary substructure
\begin{align*}
M \prec L_\theta
\end{align*}
such that $A \subseteq M$ and $|M|^L = \kappa$.
Let
\begin{align*}
\pi : M \to N
\end{align*}
be the Mostowski collapse map, where $N$ is the transitive collapse of $M$. Since $M \prec L_\theta$ and $L_\theta$ is a level of the constructible hierarchy, the Condensation Lemma for $L$ (citing a result not yet in the wiki: Condensation Lemma for constructible levels) gives an ordinal $\beta$ such that
\begin{align*}
N = L_\beta.
\end{align*}
Because $|M|^L=\kappa$, also $|L_\beta|^L=\kappa$. In $L$, every ordinal $\alpha \leq \beta$ belongs to $L_\beta$, so the ordinal $\beta$ injects into $L_\beta$. Therefore $|\beta|^L \leq \kappa$, and by the definition of $\kappa^+$ we have
\begin{align*}
\beta < \kappa^+.
\end{align*}
Since $\kappa+1 \subseteq M$, the collapse map fixes every ordinal $\alpha \leq \kappa$. Because $X \subseteq \kappa$ and $X \in M$, it follows that
\begin{align*}
\pi(X) = X.
\end{align*}
Hence $X \in N = L_\beta \subseteq L_{\kappa^+}$. Since $X \in \mathcal{P}_L(\kappa)$ was arbitrary,
\begin{align*}
\mathcal{P}_L(\kappa) \subseteq L_{\kappa^+}.
\end{align*}
[guided]
The goal is to prove that no subset of $\kappa$ first appears at or above stage $\kappa^+$ of the constructible hierarchy. Fix a constructible subset $X \subseteq \kappa$. Since $X \in L$, there is some ordinal $\theta$ with $X \in L_\theta$. We choose $\theta$ larger than $\kappa^+$ and large enough that $L_\theta$ has the finite amount of set-theoretic structure needed for elementary submodels and condensation.
Now define the parameter set
\begin{align*}
A := \{X\} \cup (\kappa+1).
\end{align*}
We include $X$ itself as an element of $A$, not merely the members of $X$, because later we must apply the collapse map to $X$. This set has $L$-cardinality $\kappa$: the ordinal $\kappa+1$ has cardinality $\kappa$ because $\kappa$ is infinite, and adjoining the single element $X$ does not increase that cardinality. By the Downward Löwenheim-Skolem theorem inside $L$ (citing a result not yet in the wiki: Downward Löwenheim-Skolem theorem), applied to the structure $(L_\theta,\in)$ and the parameter set $A$, there is an elementary substructure
\begin{align*}
M \prec L_\theta
\end{align*}
such that $A \subseteq M$ and $|M|^L=\kappa$.
Because $A \subseteq M$, we have both $X \in M$ and $\kappa+1 \subseteq M$. The reason for putting $\kappa+1$ into $M$ is that the collapse must fix every ordinal below or equal to $\kappa$. Let
\begin{align*}
\pi : M \to N
\end{align*}
be the Mostowski collapse map. The structure $N$ is transitive, and $\pi$ is an isomorphism from $(M,\in)$ onto $(N,\in)$. Since $M$ is elementary in a constructible level $L_\theta$, the Condensation Lemma for $L$ (citing a result not yet in the wiki: Condensation Lemma for constructible levels) gives an ordinal $\beta$ with
\begin{align*}
N = L_\beta.
\end{align*}
We next check that this $\beta$ lies below $\kappa^+$. Since $M$ has $L$-cardinality $\kappa$ and $\pi$ is a bijection from $M$ onto $L_\beta$, we have $|L_\beta|^L=\kappa$. The ordinal $\beta$ injects into $L_\beta$ in $L$, because every ordinal below $\beta$ is an element of $L_\beta$. Therefore $|\beta|^L \leq \kappa$. Since $\kappa^+$ is the least $L$-cardinal larger than $\kappa$, this implies
\begin{align*}
\beta < \kappa^+.
\end{align*}
Finally, because $\kappa+1 \subseteq M$, the Mostowski collapse fixes every ordinal $\alpha \leq \kappa$. Since $X \subseteq \kappa$ and $X \in M$, every member of $X$ is fixed, so the image of $X$ under the collapse is exactly $X$:
\begin{align*}
\pi(X) = X.
\end{align*}
But $\pi(X) \in N = L_\beta$, hence $X \in L_\beta$. Since $\beta < \kappa^+$, we get $L_\beta \subseteq L_{\kappa^+}$, and therefore $X \in L_{\kappa^+}$. As $X$ was arbitrary, this proves
\begin{align*}
\mathcal{P}_L(\kappa) \subseteq L_{\kappa^+}.
\end{align*}
[/guided]
[/step]
[step:Count the subsets of $\kappa$ using the size of $L_{\kappa^+}$]
We use the standard cardinal estimate for constructible levels (citing a result not yet in the wiki: cardinality of constructible levels): for every infinite ordinal $\delta$,
\begin{align*}
|L_\delta|^L \leq |\delta|^L.
\end{align*}
Applying this with $\delta=\kappa^+$ gives
\begin{align*}
|L_{\kappa^+}|^L \leq \kappa^+.
\end{align*}
Since $\kappa^+ \subseteq L_{\kappa^+}$, the reverse inequality also holds, so
\begin{align*}
|L_{\kappa^+}|^L = \kappa^+.
\end{align*}
From the inclusion proved above,
\begin{align*}
\mathcal{P}_L(\kappa) \subseteq L_{\kappa^+},
\end{align*}
we obtain
\begin{align*}
|\mathcal{P}_L(\kappa)|^L \leq |L_{\kappa^+}|^L = \kappa^+.
\end{align*}
[/step]
[step:Conclude the Generalised Continuum Hypothesis in $L$]
Combining the lower and upper bounds gives
\begin{align*}
\kappa^+ \leq |\mathcal{P}_L(\kappa)|^L \leq \kappa^+.
\end{align*}
Therefore
\begin{align*}
|\mathcal{P}_L(\kappa)|^L = \kappa^+.
\end{align*}
Since $\mathcal{P}_L(\kappa)$ is exactly the power set of $\kappa$ computed in $L$, this says
\begin{align*}
(2^\kappa)^L = (\kappa^+)^L.
\end{align*}
The cardinal $\kappa$ was an arbitrary infinite cardinal of $L$, so $(L,\in)$ satisfies the Generalised Continuum Hypothesis.
[/step]
