$\mathcal P(X) \cong \mathcal P(X)$ without singletons if the set $X$ is infinite I did not find this question here though I'm sure that it has already been asked. Sorry for that.
I am looking for an elementary way to show the following: Let $X$ be an infinite set, then there is a bijection
\begin{equation}
\mathcal P(X) \longrightarrow \{Y \in \mathcal P(X) \mid \#Y \neq 1 \},
\end{equation}
i.e. the power set $\mathcal P(X)$ does not change its cardinality when subtracting a set of cardinality $\#X$, the singletons in this case.

Take an infinite injective sequence $x_n \in X$, $n \geq 1$. 
Define $i_n : X \rightarrow X \backslash \{x_1, \ldots, x_{n-1}\}$ by having $i_n(x_k)=x_{k+n-1}$ and $i_n(y)=y$ if $y$ is not in the sequence. 
Define $j_n: X \rightarrow \{F \subset X,\, |F|=n\}$ by $j_n(z)=\{x_1, \ldots, x_{n-1},i_n(z)\}$. 
Note that each $j_n$ is well-defined and injective; that $j_1$ is surjective; that the images of the $j_n$ are pairwise disjoint. 
Now define the bijection $f: \mathcal{P}(X) \rightarrow \{F \subset X,\, |F| \neq 1\}$ as follows: if $F \subset X$ is not some $j_k(y)$ then $f(F)=F$, else, if $F=j_k(y)$, then $(k,y)$ is well-defined and we set $f(F)=j_{k+1}(y)$. 
