$\aleph(X) < \aleph(\mathcal{P}(\mathcal{P}(\mathcal{P}(X))))$ Prove: $\aleph(X) < \aleph(\mathcal{P}(\mathcal{P}(\mathcal{P}(X))))$
With $W(X)=\{\langle A,R\rangle: A \in \mathcal{P} (X),R \in \mathcal{P}(X \times X)$ and $ R $ wellorders $ A \}$ 
And $\aleph(X) =$ img $\{\langle x , \alpha \rangle: x \in W (X) , \alpha$ is an ordinal number and $\alpha$ is the type of $x \}$.
I made an injective function:
$f:\aleph(X) \to \aleph(\mathcal{P}(\mathcal{P}(\mathcal{P}(X)))): \alpha \mapsto \{\{\{\alpha\}\}\}$
Can anyone give the full prove?
 A: Your function is not really an injection. What you essentially need to show is that there is an injection from $\aleph(X)$ into $\mathcal{P(P(P}(X)))$, which will conclude that its $\aleph$ is larger than that of $X$ itself.
Recall that $\aleph(X)$ is the least ordinal $\kappa$ such that there is no injection from $\kappa$ into $X$. Therefore if $\alpha<\aleph(X)$ then there is an injection from $\alpha$ into $X$.
This means that there are subsets of $X$ which can be well-ordered of order type $\alpha$. But alas, given a countably infinite subset there are many ways to well-order it (even in the same order type!).
But if $A$ is a subset of $X$ which can be well-ordered of type $\alpha$, then there exists a chain in $\mathcal P(X)$ witnessing that. Namely, if $(A,<)$ is of order type $\alpha$ and $f$ is an isomorphism, then the chain $\Big\{\{a\in A\mid f(a)<\beta\}\mid\beta<\alpha\Big\}$ is a chain in $\mathcal P(X)$ whose order type, when ordered by $\subseteq$, is $\alpha$.
As before, if there is one, then there are many. Now we can consider a $C_\alpha=\{\mathcal A\subseteq\mathcal P(X)\mid (\mathcal A,\subseteq)\cong(\alpha,\in)\}$ as a subset of $\mathcal{P(P(}X))$, which includes all the chains of order type $\alpha$. This set is unique, meaning that now we have taken all the possible ways to construct a chain of order type $\alpha$ in $\mathcal P(X)$. So we can injectively identify where to send $\alpha$.
Therefore the function $\alpha\mapsto C_\alpha$ is an injection from $\aleph(X)$ into $\mathcal{P(P(P(}X)))$. Therefore $\aleph(X)<\aleph(\mathcal{P(P(P(}X))))$.
