Do six such numbers exist? 
Is it possible to find $6$ integers $a_1,a_2,\ldots,a_6 \geq 2$ such that $$a_1+a_1a_2+a_1a_2a_3+a_1a_2a_3a_4+a_1a_2a_3a_4a_5+a_1a_2a_3a_4a_5a_6 = 248?$$

I was wondering how we could establish the existence of such numbers. Is there a way to do it without finding the actual $6$ numbers?
 A: Note that:$$a_1(1+a_2(1+a_3(1+a_4(1+a_5(1+a_6))))) = 248 = 2^3(31)$$ 
$a_1$ has to be 2, because $1+a_2(1+a_3(1+a_4(1+a_5(1+a_6))))\geq 63.$ And so now we're solving for:
$$a_2(1+a_3(1+a_4(1+a_5(1+a_6)))) = 123.$$ 
Following the same kind of argument, keeping in mind the constraint $a_i\geq2$, we arrive to the conclusion that there do not exist such numbers.
A: I will provide the first step, from which you can hopefully complete the proof yourself. As @snulty mentioned, we can factor this as follows:
$$
a_1(1 + a_2(1 + a_3(1 + a_4(1 + a_5(1 + a_6))))) = 2^3 \cdot 31.
$$
This implies that $a_1 = 2$ or $a_1 \geq 4$. However, if $a_1 \geq 4$, we have 
$$
a_1(1 + a_2(1 + a_3(1 + a_4(1 + a_5(1 + a_6))))) \geq 4(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2))))) = 252,
$$
hence this is not an option. 
We find $a_1 = 2$, hence the problem is reduced to
$$a_2(1 + a_3(1 + a_4(1 + a_5(1 + a_6)))) = 123.$$
The prime factorization of $123$ is $3 \cdot 41$, so what can we conclude about the value of $a_2$? Finish the rest of the proof!
