function from N\mathbb{N} to N×N×N\mathbb{N}\times\mathbb{N}\times\mathbb{N}

Here is one way: I assume $0\in{\mathbb N}$. Given a triple $(a,b,c)$, assign to it the number $2^a(2(2^b(2c+1)-1)+1)-1$. The same idea shows that ${\mathbb N}$ has the same size as ${\mathbb N}^k$ for any positive integer $k$.

(This can be easily modified and in fact gives a slightly easier formula if your convention is that the naturals begin with 1.)

There are of course many other ways. The idea I am using above is to first find a bijection between ${\mathbb N}\times{\mathbb N}$ and ${\mathbb N}$, and then use it to create the one you want. The bijection I used is $(a,b)\mapsto 2^a(2b+1)-1$. Another popular choice uses Cantor’s pairing function, $\displaystyle (a,b)\mapsto \frac{(a+b)(a+b+1)}2 +b$. As a technical comment, there is actually an advantage to Cantor’s choice, in that it is a polynomial, and it is easier to define (in a first-order way) in the usual structure of the natural numbers $({\mathbb N},+,\times,0,1,<)$. The choice I used requires exponentiation, and although this can also be defined in a first-order way, it is significantly less trivial.

That exponentiation is first-order definable is known since Gödel's work on the incompleteness theorem. In fact, there is a way to define any recursive function. For exponentiation, I sketch the details in this answer. The bulk of the argument for Hilbert's 10th problem is that in fact there is a polynomial definition of exponentiation, in the sense mentioned here. This is significantly more involved, and was rather surprising at the time.