An analogue to Cantor's theorem Cantor's theorem states that for all sets $$|A| < |2^A|$$
I was interested in a similar proposition. If $A$ is a set, denote by $A! := \{f : A \rightarrow A \mid f \text{ is a bijection}\}$. Is it true in general that $|A| < |A!|$?
It is not to difficult to show that $|\mathbb{N}| < |\mathbb{N}!|$. To see this, let $B \subset \mathbb{N!}$ be the set of permutations which are either the identity or have some number of even naturals swapped with their right neighbor. (0 1 (3 2) 4 5 (7 6) 8 9 10 11 12 ... is an example). Then $B$ is uncountable because if we map unswapped pairs to 0 and swapped pairs to 1, this constitutes a bijective mapping into the set of infinite binary strings, which we know to be uncountable by the classic diagonalization argument. Then as $B \subset \mathbb N!$, $\mathbb N!$ is uncountable.  
Unfortunately this proof does not yield an approach to the general case. Any ideas?

Taking a hint from Unit’s comment down below (ultimately coming from Factorials of Infinite Cardinals by Dawson and Howard, it seems), here’s a case differentiation proving the result in ZF (as far as I can tell).

In case $|A| = |2 × A|$: Mapping $2^A → (2×A)!,~ B ↦ σ_B$, where $σ_B$ swaps the two copies of $B$ in $2×A  = A \sqcup A$ pointwise and fixes the rest, is injective. If $2×A \cong A$ you have $(2×A)! \cong A!$ and hence this proves $|A| < |2^A| ≤ |A!|$.
In case $|A| < |2 × A|$: Assume $|A| > 2$ and fix a two element set $2_A ⊂ A$ and a check-point $ξ ∈ A\setminus2_A$. Then the mapping $2_A×A → A!,~(α,x) ↦ (α~ξ~x)$ is evidently left-inverted by $A! → 2_A×A,~τ ↦ (τ^{-1}(ξ),τ(ξ))$, hence it’s injective, proving $|A| < |2×A| ≤ |A!|$.

I hope this is correct.
