$$\bigcup\limits_{n\in\mathbb N} A_n$$
The book is asking me to prove that $f(\bigcup\limits_{n\in\mathbb N} A_n) = \bigcup\limits_{n\in\mathbb N} A_n$.
I'm able to prove that f(the notation above) = the notation above U A(n+1) (A(n+1) looks like An above) does that mean that f(the notation above) = the notation above? since n+1 is in N
Does this notation mean the union of the sets $A_1 \cup A_2 \cup ... \cup A_{6565656}$?
On
$$\bigcup_{n \in \Bbb{N}}A_{n} = A_{1} \cup A_{2} \cup A_{3} \cup \ .....$$ Next you are trying to show that $$f\left(\bigcup_{n \in \Bbb{N}}A_{n}\right) = \bigcup_{n \in \Bbb{N}}f\left(A_{n}\right)$$ These types of proofs are often done by double containment. Let $x \in f\left(\bigcup_{n \in \Bbb{N}}A_{n}\right)$. Then $f^{-1}(x) \in \bigcup_{n \in \Bbb{N}}A_{n}$, so there exists at least one $i \in \Bbb{N}$ where $f^{-1}(x) \in A_{i}$. But this means $x \in f(A_{i})$, so obviously $x \in \bigcup_{n \in \Bbb{N}}f\left(A_{n}\right)$. This establishes that $$f\left(\bigcup_{n \in \Bbb{N}}A_{n}\right) \subseteq \bigcup_{n \in \Bbb{N}}f\left(A_{n}\right).$$
Now let $y \in \bigcup_{n \in \Bbb{N}}f\left(A_{n}\right)$. Again, there has to be some $j \in \Bbb{N}$ where $y \in f(A_{j})$, which means $f^{-1}(y) \in A_{j}$. But if $f^{-1}(y) \in A_{j}$ then $f^{-1}(y) \in \bigcup_{n \in \Bbb{N}}A_{n}$. Now we apply $f$ again to both sides of this subset relation and we have $y \in f\left(\bigcup_{n \in \Bbb{N}}A_{n}\right)$. Now we have established that $$ \bigcup_{n \in \Bbb{N}}f\left(A_{n}\right) \subseteq f\left(\bigcup_{n \in \Bbb{N}}A_{n}\right).$$ By double containment, we may conclude that $$f\left(\bigcup_{n \in \Bbb{N}}A_{n}\right) = \bigcup_{n \in \Bbb{N}}f\left(A_{n}\right)$$
Remember that the natural numbers are denoted by $\mathbb{N} = \{1,2,3,\ldots\}$.
And yes, you are correct:
$$\bigcup_{n \in \mathbb{N}} A_n = A_1 \cup A_2 \cup A_3 \cup \ldots $$
The big union symbol is just notation to express taking lots of unions more easily.
We sometimes refer to unions like $\bigcup_{n \in \mathbb{N}} A_n$ as infinite unions, because they take the union of an infinite (but countable) number of sets.