If $\sum\limits_{i=1}^na_i=\prod\limits_{i=1}^na_i$ for every $n$, identify $\lim\limits_{n\to \infty}a_n$ 
Let $\left(a_n\right)_{n \in\mathbb{N}} $ denote a sequence of real numbers such that, for every $n\geqslant1$, $$\sum_{i=1}^na_i=\prod_{i=1}^na_i$$ Identify the limit $$\lim_{n\to \infty}a_n$$

What I have done: $$a_1-a_1=0 \\a_1+a_2-(a_1a_2)=0 \to  a_1\cdot a_2(\dfrac{a_1}{a_2}+\dfrac{a_2}{a_1}-1)=0\\.\\.\\.\\a_1\cdot a_2\cdot \cdot \cdot a_n(\dfrac{a_1}{a_2\cdot \cdot \cdot a_n}+\dfrac{a_2}{a_1\cdot \cdot \cdot a_n}+...+\dfrac{a_n}{a_2\cdot \cdot \cdot a_{n-1}}-1)=0$$
Now what do I do ?
 A: The answer is: if $a_1 > 1$, then the limit is 1; if $a_1 < 1$, then the limit is 0; and there is no sequence with $a_1 = 1$.
First, notice that if we let $S_n = \sum_{k=1}^n a_k$, then
$$S_n + a_{n+1} = \left( \sum_{k=1}^n a_k \right) + a_{n+1} = \sum_{k=1}^{n+1} a_k = \prod_{k=1}^{n+1} a_k = \left( \prod_{k=1}^n a_k \right) a_{n+1} = S_n a_{n+1}.$$
Solving $S_n + a_{n+1} = S_n a_{n+1}$ for $a_{n+1}$ gives $a_{n+1} = \frac{S_n}{S_n - 1}$ (and also implies that $a_1 = S_1 \ne 1$).
Now for the first case, suppose $a_1 > 1$.  Then $a_{n+1} = \frac{S_n}{S_n - 1}$, so by induction, we can conclude that $a_n > 1$ and $S_n > 1$ for every $n$.  Thus, $S_n = \sum_{k=1}^n a_k > n$, and so $a_{n+1} = 1 + \frac{1}{S_n - 1} < 1 + \frac{1}{n-1}$ for $n > 1$.  By the squeeze theorem, we can conclude $a_n \to 1$ as $n \to \infty$.
Otherwise, suppose $a_1 < 1$.  Then we get $S_{n+1} = S_n + \frac{S_n}{S_n - 1} = \frac{S_n^2}{S_n - 1}$.  Now we can conclude by induction that $S_n \le 0$ for $n \ge 2$, and $S_n$ is nondecreasing for $n \ge 3$ since $a_n = \frac{S_{n-1}}{S_{n-1}-1} \ge 0$.  It follows that $S_n$ must have some limit $L \le 0$, which must then satisfy $L = \frac{L^2}{L-1}$.  This forces $L = 0$, so $a_n = S_n - S_{n-1} \to 0 - 0 = 0$ as $n \to \infty$.
