Meaning of the set $\mathbb N^\mathbb N$ I came across a question which requires one to check if there's a bijection from the set $ \mathbb N^\mathbb N$ to another set. I've never seen a set defined this way and was wondering if this was just a typo. Could anyone clarify this for me? If this is correct, then what does the set actually mean? I mean, I can think of elements in $\mathbb R$ or in $\mathbb N$, but how do I think of elements in $\mathbb N^\mathbb N$?
 A: The elements of $A^B$ are function of the form $f\colon B\to A$. That is function whose domain is exactly $B$ and their values are elements from $A$.
If so, $\Bbb{N^N}$ is the set of all functions from $\Bbb N$ to itself. Or the set of infinite sequences of natural numbers (as a sequence is just a function).

As a footnote remark, I should add that this set is sometimes denoted by ${}^BA$ instead of $A^B$, especially in contexts were exponentiation is used in set theory.
A: To elaborate on / illustrate Asaf's answer, and to provide some context for why this notation makes sense:
Let $A$ be the set of letters in the alphabet $\{a,b,\dots,z\}$. Let $B$ be the set of digits $\{0,\dots,9\}$. Then $A^B$, the set of functions from $B$ to $A$, is the set of all ways I could assign a letter to each digit. For example, one element of $A^B$ is the function
$$0\mapsto a$$
$$1\mapsto b$$
$$2\mapsto c$$
$$\vdots$$
$$9\mapsto j$$
Another element of $A^B$ is the function
$$0\mapsto z$$
$$1\mapsto y$$
$$2\mapsto x$$
$$\vdots$$
$$9\mapsto q$$
Yet another is the function
$$0\mapsto m$$
$$1\mapsto m$$
$$2\mapsto m$$
$$\vdots$$
$$9\mapsto m$$
Of course there are a lot of other choices. All my choices had nice patterns that made them easy to write down, but most functions are much more random-seeming in terms of which digit gets which letter. In fact, each function can be constructed by going through the digits and choosing one of the 26 letters for each of them. There are a lot of choices, so the set of functions $A^B$ is big.
Just how big?
Well, what choices have to get made to determine a function? For each of the 10 digits, you have to choose one of the $26$ letters. Thus you choose 10 times among 26 choices. Therefore the total number of functions is $26^{10}$.
Notice that $26^{10}=(\text{size of }A)^{\text{size of B}}$. This is why the notation $A^B$ is used! Because that way, we get
$$\text{size of }A^B = (\text{size of }A)^{\text{size of B}}$$
Back to your original question, $\mathbb{N}^\mathbb{N}$ refers to the set of functions from $\mathbb{N}$ to $\mathbb{N}$, just as in the above toy example. As Asaf mentioned, this can also be looked at as the set of infinite sequences of natural numbers, since a sequence like $1,1,2,5,14,\dots$ can be seen as a function that takes a position in the sequence to the entry found in that spot:
$$1\mapsto 1$$
$$2\mapsto 1$$
$$3\mapsto 2$$
$$4\mapsto 5$$
$$5\mapsto 14$$
$$\vdots$$
