Nested... binomials coefficients? Can I have a proof that this number exists?
The number:
$$\binom{1}{\binom{2}{\binom{3}{\binom{4}{\vdots}}}}$$
If the number exists, then what is the closed form of that number?

Note that ${i\choose i+1} = 0$. 
So $A_2 ={1 \choose 2} = 0$,
$A_3 ={1\choose{2\choose 3}} = {1 \choose 0} = 1$
Now $A_4={1\choose{2\choose {3\choose 4}}} = {1\choose{2\choose 0}} = {1\choose{1}} = 1$
Lastly
$$A_5={1\choose{2\choose {3\choose {4\choose 5}}}} = {1\choose{2\choose {3 \choose 0}}} = {1\choose{2\choose 1}} = {1\choose 2} = 0$$
If you consider a translation
$${1+1\choose{2+1\choose {3+1\choose {4+1\choose 5+1}}}} = {1+1\choose{2+1\choose {3+1 \choose 0}}} = {1+1\choose{2+1\choose 1}} = {1+1\choose 2+1} = 0$$
So you find a period for your expression.
Consider $$A_n = {1 \choose {2 \choose \underset{n}{\vdots} }}$$
Since $${k+1\choose{k+2\choose {k+3\choose {k+4\choose k+5}}}} = \ldots   = 0 = {k+1\choose k+2}$$ 
We obtain $A_{10} = A_7 = A_4 = 1 $. To the general case take $n \mod 3 = r$ $A_n = A_{r + 3}$ or you could consider $A_0 = 1A_1 = 1$ then $$A_0 = A_3,A_1 = A_4, A_2= A_5 $$
That is $A_n = A_r$ (where $r = n \mod 3$)
