Let $n\in \mathbb{N}$ be a cardinal distinct from zero.
How to prove in ZFC that there is not set containing all sets of $n$ elements?
$\textbf{Edit 1.}$
If I suppose that the is such a set that contains all sets of $n$ elements, in particular, for $n = 1$ it is the unitary set of all sets. But that set does not exists in ZFC, so that would be a contradiction?
Let $A_n$ stand for the class of sets with $n$ elements. In case of $n=1$, we can use $A_1\in\{A_1\}\in A_1$ to build an infinite chain of membership relationship, which contradicts well-foundedness.
More generally, we have $$A_n\in\{A_n, \mathcal P(A_n), \mathcal P(\mathcal P(A_n)), \cdots, \mathcal P^{n-1}(A_n)\}\in A_n$$ where $A_n, \mathcal P(A_n), \cdots$ are all distinct due to cardinality.
For a fixed $n$ let $A_n$ be the collection of all $n$-element sets. By the Axiom of Replacement if you have any set and any class function defined on that set then the image of the function is again a set. Thus if you can find a class function defined on $A_n$ whose image is something you already know to be a proper class, then $A_n$ is also a proper class.
In the comments the case $n=1$ was addressed: The class function $F : A_1 \to V$ defined by $F(x) = \bigcup x$ (or, equivalently, $F(\{a\}) = a$) is surjective, and $V$ is not a set, so $A_1$ cannot be a set.
For higher values of $n$ you can do similar kinds of things. For $n=2$ the function $F : A_2 \to V$ defined by $F(x) = \bigcup x$ is almost surjective - it just misses $\emptyset$. But you (probably?) already know that $V \setminus \{\emptyset\}$ is also a proper class. At $n=3$ the same definition is produces a surjection from $A_3$ to $V \setminus (\{\emptyset\} \cup A_1)$, which is again a proper class (because we already saw that $A_1$ is a set!). And so forth...
Chris Eagle’s answer is absolutely correct, but I think there’s a different, valuable way of looking at this. Specifically, I think this is best viewed as a special case of a more general tool. I’ll write it up as a separate answer.
To fix a little terminology and notation, let a “class” denote any collection of sets, and let $V$ denote the class of all sets. I am assuming you’re already familiar with the formal reasons that $V$ is not a set.
There are loads of different sufficient conditions for showing that a class is not a set. Here’s one of my favorites:
Theorem: Let $\mathcal{C}$ be a class. If for all sets $S \in V$, there is a an element $C \in \mathcal{C}$ such that $S \in C$, then $\mathcal{C}$ is not a set.
Proof: To prove this theorem, recall the axiom of union, which says: $$\forall F \exists A \forall Y \forall x \big [(x \in Y \wedge Y \in F) \rightarrow x \in A \big ] $$ Suppose that $\mathcal{C}$ is a set. Then we may apply this axiom to $\mathcal{C}$ and find that there is a set $A$ such that for all sets $Y \in \mathcal{C}$ and all $x \in Y$, we have that $x \in A$. By our condition on $\mathcal{C}$, we know that for every set $S$, there is a $C \in \mathcal{C}$ such that $S \in C$. It follows that for every set $S$, we have that $S \in A$. Thus $V \subseteq A$.
Additionally, we have that $A$ is a set, so we have that $A$ is a set containing all sets. Since the formula $x=x$ picks out all sets, it follows by restricted comprehension that $\{x \in A : x = x\} = V$ is a set. But there is no set of all sets. Thus $\mathcal{C}$ could not have been a set to begin with. □
This is a quite “general test” for checking if a class is not a set. We may easily apply it to your example.
Corollary: For $n > 0$, let $F_n$ denote the class of sets with exactly $n$ elements. $F_n$ is not a set.
Proof: There are $n$ distinct sets. Thus for any set $S$, there is a set $T$ such that $S \in T$ and $T$ has exactly $n$ elements. Thus we meet the hypotheses of the theorem. □
