As others have mentioned, it's typical to define
$$(a_1,\ldots, a_n)=((a_1,a_2,\ldots, a_{n-1}),a_n)$$
though I like to call them "ordered $n$-tuplets" to distinguish from the alternative definition I give below (whose naming convention is rather standard when used for infinite sets, in which the above definition cannot be extended).
I think it's worth noting that there are a lot of different ways to extend the notion of an ordered-pair. For example, once ordered-pairs and ordered-triplets have been defined, you can define $A\times B$ for two sets $A$ and $B$ and a function as an ordered-triple $(A,B,f)$, where $f\subset A\times B$ is well-defined and left-total. For there, you can then "redefine" $n$-tuples to be surjective maps $f: \{0,\ldots, n-1\}\rightarrow A$. (Strictly speaking, we could get away with only having ordered-pairs if we want to just define an $n$-tuple to be a subset of $\{0,\ldots, n-1\}\times A$ for some set $A$ that is well-defined and left-total. But defining functions as triples is useful because it allows one to talk about surjectivity.)
Another approach is defining $(a,b):=\{a,\{a,b\}\}$, though that this satisfies the property that $(a,b)=(c,d)$ implies $a=c,b=d$ requires the Axiom of Foundation.
For the sake of differentiating between the ordered $n$-tuplets as defined above and these $n$-tuplets, I'll denote an $n$-tuple as $\langle a_0,\ldots, a_{n-1}\rangle$ where $a_0,\ldots, a_{n-1}$ are the images of the function $f$ on $0,\ldots, n-1$, respectively. The nice property that $n$-tuples have that $n$-tuplets don't have is that given an $n$-tuple $\mathbf{a}=\langle a_0,\ldots, a_{n-1}\rangle$ and an $m$-tuple $\mathbf{b}=\langle b_0,\ldots, b_{m-1}\rangle$, we have $\mathbf{a}=\mathbf{b}$ if and only if $n=m$ and $a_i=b_i$ for each $i\in\{0,\ldots, n-1\}=\{0,\ldots, m-1\}$, the reason being that two functions being equal implies their domains are equal.
As mentioned above, when we want to extend the notion of an $n$-tuple(t) to deal with indexing things by arbitrary sets, we have to use the notion of an $n$-tuple, defining an $I$-tuple (or a family of sets indexed by $I$) to be a surjective map $f:I\rightarrow A$. If want to regard these $I$-tuples as simply being elements of some infinite Cartesian product $\prod_{i\in I}{A_i}$ (where $(A_i)_{i\in I}$ is a family of sets indexed by $I$), we could choose the elements to either be indexed families $f:I\rightarrow A$ where $A\subset \bigcup_{i\in I}{A_i}$ and $f(i)\in A_i$ for each $i\in I$ (so $f$ is necessarily surjective by definition), or we could choose to take the elements to be simply functions $f:I\rightarrow \bigcup_{i\in I}{A_i}$ where $f(i)\in A_i$ for each $i\in I$. Restricting to the finite case where $I=\{0,\ldots, n-1\}$, we then have a few different ways we could have defined, say, $A\times B$ (though we did use the Kuratowski definition to define a function in the first place, nothing stops us from then considering as "ordered pairs" $2$-tuples rather than $2$-tuplets from now on).
Given these alternatives exist for definitions of $n$-tuple(t)s or infinite cartesian products, a question might be which is better. Ultimately, it's kind of an arbitrary choice, in the sense that there is a natural way to switch between the two that "preserves" how the canonical projections act on the Cartesian products.
Best Answer
You seem to be quoting an ill-stated version of the definition of a relation. A better crafted version will have the domain and range given as part of the definition: