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
We accept this definition because it works, and it works very well.
However do note that almost nobody cares about the actual encoding of ordered pairs, and in most cases set theorists don't really care either. You just want a definition which satisfies the property which you have quoted.
I can't recall any substantial proof in set theory which actually refer to this definition of ordered pairs. Instead we just use the fact that there is some $\varphi(x,y,z)$ which is true if and only if $z$ represents the ordered pair $(x,y)$. Had I chose a different encoding, the proof would stay the same.
So why do we use it? Well, it tells us that set theory is strong enough to support an internal definition of ordered pairs which is another "thumbs up" in its favor as a foundational theory. But also we use it because it was there when we needed such definition, and once you have a working machine you don't really bother replacing the cogs which work perfectly.
Note that on its own, the object of an ordered pair is very uninteresting. So we don't really bother with finding a "better" definition, because nobody really cares about how you encode. We just want to know that it exists within our mathematical universe, and if our mathematical universe is based on set theory (in one way or another) then we want to know that we don't need an extra type in the world in order to have ordered pairs.