Q:

How to extend definition of n-tuple to the case $n=0$?

The classical definition of n-tuple $(x_i)_{i < n}$ starts at $n=2$. In this case $$(x_0,x_1) := \{\{x_0\},\{x_0,x_1\}\}$$(1).
For $2<n=k+1$, $(x_i)_{i < n}:=((x_i)_{i < k},x_k)=\{\{(x_i)_{i < k}\},\{(x_i)_{i < k},x_k\}\}$(2). 
Hence it well defined on $2 \le n \in \mathbb N$.
One would tend to extend this definition to every natural number. 
In the case $n=1$, if we let $(x_0)=x_0$, then (2) can be extended to $1 \le n \in \mathbb N$, so $(x_0,x_1)=\{\{(x_0)\},\{(x_0),x_1\}\}=\{\{x_0\},\{x_0,x_1\}\}$, which also compatible with (1).  
However, if we go on do this to $n=0$, we will get a trouble.
That is, $()=?$ If $()=\emptyset$, then $(x_0)=((),x_0)=\{\{()\},\{(),x_0\}\}=\{\{\emptyset\},\{\emptyset,x_0\}\}$, but it contradict to $(x_0)=x_0$ as defined. Similar, if we let $()$ remains blank, then $(x_0)=((),x_0)=\{\{()\},\{(),x_0\}\}=\{\{\},\{ ,x_0\}\}=\{\emptyset,\{x_0\}\}$, which also contradict to $(x_0)=x_0$.
So my question: Is there any methodology to climb over this barrier?

A:

There is no way to define a $0$-tuple to that $1$-tuples are ordered pairs (without contradicting Foundation):  Suppose that there were such a definition of a $0$-tuple $\langle\;\rangle$ so that $$a = \langle a \rangle = \langle \langle \; \rangle , a \rangle = \{ \{ \langle \; \rangle \} , \{ \langle \; \rangle , a \} \}$$  then $a \in \{ \langle \; \rangle , a \} \in a$.

