Let $A$ be infinite and $\operatorname{Sq}(A)$ be the set of finite sequences in $A$. Then $A$ and $\operatorname{Sq}(A)$ are equinumerous.
My attempt:
Let $A_n$ be the set of finite sequences of length $n$. Then $A_n=A^n$ where $A^n$ is the $n$-ary Cartesian product over $A$.
Assuming Axiom of Choice, we have $|X|=|X\times X|=|X^2|$ for an infinite set $X$. By induction on $n$, we can prove that $|X^n|=|X|$ for all $n\in \Bbb N$. As a result, we have $|A_n|=|A^n|=|A|$ for all $n\in\Bbb N$.
Assuming Axiom of Choice and by induction on $n$, we also have $|\bigcup_{n\in\Bbb N}X_n|=|X|$ if $|X_n|=|X_m|\ge\aleph_0$ and $X_n\cap X_m=\emptyset$ for all $n,m\in\Bbb N$ and $n\neq m$.
Finally, $|\operatorname{Sq}(A)|=|\bigcup_{n\in\Bbb N}A_n|$ where $|A_n|=|A_m|=|A|\ge\aleph_0$ and $A_n\cap A_m=\emptyset$ for all $n,m\in\Bbb N$ and $n\neq m$. Thus $|\operatorname{Sq}(A)|=|\bigcup_{n\in\Bbb N}A_n|=|A|$ and hence $A$ and $\operatorname{Sq}(A)$ are equinumerous.
Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!