I have considered as following:
Consider any infinite set $A$ with cardinality $m$ for some infinite cardinal $m$, for each natural number $n$, denote the set of subsets of $A$ with exactly $n$ elements as $A_n$.
Denote the set of finite subsets of $A$ by $F$, thus $F=\lbrace \emptyset \rbrace \cup A_1\cup A_2 \cup...=\lbrace \emptyset\rbrace \oplus A_1\oplus A_2\oplus...$.
There is the missing argument to conclude that for each $n\in \Bbb N$, $A_n\approx A$
Hence $\#F=\#(\lbrace \emptyset\rbrace \oplus A_1\oplus A_2\oplus...)=1+m+m+... =m+m+...=(1+1+1+...)m=\aleph_0\cdot m=m$
Is that correct? I fell I am stuck on proving that each $A_n$ has the same cardinality as $A$. It seems to be related to the product set but I can only find an injection from each $A_n$ to $A^n$. So could someone please help? Thanks so much!
The collection $A_k$ of $k$-element subsets of $A$ has cardinality at most $|A^k|$, since a subset $A^{(k)}$ of $A^k$ maps onto $A_k$ by $(a_1,\ldots,a_k)\mapsto\{a_1,\ldots,a_k\}$. ($A^{(k)}$ will be the collection of $k$-tuples of distinct elements) By standard cardinal arithmetic $|A^k|=|A|^k =|A|$ when $A$ is infinite. Therefore $|A_k|\le |A|$. That's all you need, equality is true, but $\le$ is good enough.