Vtyjkyuk

Let $$mathrm$$ as $A$ is a given set such that $|A| > 1$.

For every $ngeq1$, let $mathrmA^n$ be the set of function from $0,1,....,n-1$ to $mathrmA$.

prove: $$mathrm|A|=|cup_ninmathbbN^mathrm+mathrmA^n|$$

if we look on any object: $mathrmA^iinmathrmA^n$ as $iinmathbbN$ and $sum_i=1^n-1mathrmA^i=mathrmA^n$ , than any $mathrmA^i:ilongrightarrowmathrmA$ as I understand is defined to be onto, therefore $mathrmA^n$ is onto $A$, so $$|mathrmA|leq|cup_ninmathbbN^mathrm+mathrmA^n|$$
(this is just some ideas I had because I didn't really found any information about how to solve this type of Q).

after that, it is clear that I'll need to use the fact $mathrm$ in some way so I could use CBS, but i really can't find the way to solve this.

As noted in the comments, let us suppose that $A$ is neither a singleton nor empty, so that it has to be infinite (it is trivial for the empty set and false for a singleton).

The inequality $|A| leq |bigcup A^n|$ can be seen even easier by noting that there is a trivial bijection between $A$ and $A^1$.
Next, note that $A^n$ can be identified with the $n$-fold cartesian product $underbraceA times dots times A_n text times$, in particular, $A^n+1 = A^n times A$.
Using the hypothesis, you can show by induction that $|A| = |A^n|$ for every $n geq 1$. Now we construct an injection $f colon bigcup A^n rightarrow A$. Let me first explain what we want to do. Imagine $A$ as a square in the plane. Then we will embedd $A^1$ in one half of the square, embedd $A^2$ in one half of the half, i.e. a quarter, of the square, embedd $A^3$ in one eigth of the square and so on. There probably are quicker proofs but mine is explicit.

Consider the sets $A^n$ as subsets of $A^n+1$ simply by embedding them via $(a_1,dots,a_n) mapsto (a_1,dots,a_n,x)$ for some fixed element $x in A$.
Since $|A| = |A^2|$, there is a bijection $f_2 colon A_2 rightarrow A$.
Given an element in $a in A^1$ define $f(a)$ to be $f_2(a)$. This way, $f$ maps $A_1$ onto the set $f_2(A^1)$. Convince yourself that $A setminus f_2(A^1)$ still has the same cardinality as $A$ since we have $|A| = |A^2|$.
Thus, $|A setminus f_2(A^1)| = |A| = |A^3|$ and there is a bijection $f_3 colon A^3 rightarrow A setminus f_2(A^1)$. Given $a in A^2 subset A^3$ define $f(a) = f_3(a)$.
As before, $A setminus (f_2(A^1) cup f_3(A^2))$ still has the same cardinality as $A$ and we can proceed this procedure indefinitely.