Why a Grothendieck universe contains cardinalities of all its elements? Let $\mathfrak{U}$ be a Grothendieck universe; that is, a set satisfying the following properties:


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*$\forall x,y,$ $$y \in x \in \mathfrak{U} \Rightarrow y \in \mathfrak{U};$$

*$\forall x,y \in \mathfrak{U}$ $$\{x,y\} \in \mathfrak{U};$$

*$\forall x \in \mathfrak{U},$ $$\mathcal{P}(x) \in \mathfrak{U};$$

*$\forall I \in \mathfrak{U}, \forall X \subseteq \mathfrak{U}, \forall f\colon I\to X,$ $$\bigcup_{i \in I} f(i) \in \mathfrak{U};$$

*$\mathbb{N} \in \mathfrak{U}.$
What I need to know is why if $x \in \mathfrak{U},$ then necessarily $|x| \in \mathfrak{U}$. Generalliy, it's not true, of course, that if $x \in \mathfrak{U}$ and if $|y| = |x|$, then $y \in \mathfrak{U}$.
One other detail is that I would like to avoid using the fact that $\mathfrak{U} = V_\kappa$ for some inaccessible cardinal $\kappa$. Of course, if it is the only reasonable way, then...
 A: Fix a well-ordering of the set $X$.  Then by induction on this well-ordering, for every $x \in X$, the ordinal $\operatorname{rank}(x)$ corresponding to $\{ y \in X : y < x \}$ is in $\mathfrak{U}$.  Indeed, $\operatorname{rank}(x) = \bigcup_{y < x} (\operatorname{rank}(y) \cup \{ \operatorname{rank}(y) \})$.  Now, the ordinal corresponding to the order type of $X$ is $\operatorname{OT}(X) = \bigcup_{x \in X} (\operatorname{rank}(x) \cup \{ \operatorname{rank}(x) \}) \in \mathfrak{U}$ also.
Now, either $|X| \in \operatorname{OT}(X)$, or $|X| = \operatorname{OT}(X)$, and in both cases we can conclude $|X| \in \mathfrak{U}$.
A: Let $U$ be a Grothendieck universe.  $U$ is a transitive set.  If $y$ is an ordinal then $z\in y\in U\implies z\in U.$ So the set $o(U)$ of ordinals in $U$ is an ordinal. $o(U)$ is the least ordinal not in $U.$ 
For infinite $x\in U$ let $g:x\to |x|$ be a bijection. For $y\in x$ let $f(y)=g(y)$ if $g(y)<o(U)$ and let $f(y)=0$ if $g(y)\geq o(U).$ Then $\cup_{y\in x}f(y)\in U.$ 
If $|x|\geq o(U)$ then $o(U)=\cup_{y\in x}f(y)\in U,$ which is absurd because $o(U)\not \in U.$ Therefore $|x|<o(U),$ that is, $|x|\in o(U)\subset U.$ 
Of course for a finite $x\in U$ we have $|x|\in \Bbb N\subset o(U)\subset U.$
$Remarks.$ We can also see that $o(U)$ is strongly inaccessible: If $cf(o(U))<o(U)$ then $cf(o(U))\in U$ but then there is $f:cf(o(U))\to o(U)\subset U$ with $o(U)=\cup_{x\in o(U)}f(x)\in U,$ which is absurd. So $o(U)$ must be regular. And if $y< o(U)$ then (using the result of your Q),  $y\in U$ so $P(y)\in U$ so $2^y=|P(y)|\in U$ so $2^y <o(U)$. So $o(U)$ is a strong limit.   
A: (This is maybe not the cleanest way to do this one, my cardinal arithmetic is a little rusty.)
You can use the indexed union. Since $\mathbb{N}$ is in the universe (which we identify with the cardinal $\aleph_0$), take any $I$ in the universe, and define the function $f(i)=\aleph_0 \times \{i\}$. (Here we exploit the fact that if $i\in I$ then $i$ must also be in the universe.) Thus $$\bigcup_{i\in I}\aleph_0 \times \{i\}:= \sum_{i \in I} \aleph_0 = \max\{|I|,\aleph_0\} \in \mathfrak{U}. $$
I guess to get this to go through you also need to prove that Grothendieck universes are closed under Cartesian product, which is a fairly easy exercise.
(Edit: all the cardinals less than $\aleph_0$ are already in the universe, by the axiom that closes the universe under set containment.)
