Relentless reciprocals

I am not sure how to prove that the solution is unique, but if we let $x_1 = 2$ and $x_{k+1} = x_k^2-x_k+1$ for $k \ge 1$ , we get that $\frac1{x_1} + \cdots + \frac1{x_k} + \frac1{x_1 x_2 \cdots x_k} = 1$ for all $k$ . (This is an easy induction; the key is that $a_k$ is one more than the product of the previous $a_i$ .)

Mod $5$ this sequence is $2,3,2,3,2,3,\ldots$ .

Mod $7$ it's $2,3,0,1,1,1,\ldots$ .

Mod $13$ it's $2,3,7,4,0,1,1,1,\ldots$ .

So $x_{2014}$ is $3$ mod $5$ , $1$ mod $7$ , and $1$ mod $13$ . By the Chinese Remainder Theorem, it is $\fbox{183}$ mod the product.