$x_{n+1}=x_n(x_n+\frac{1}{n})$ $x_n<x_{n+1}$ for all $n$ so: $x_n+\frac{1}{n}>1$ $x_n>\frac{n-1}{n}$ $x_n<1$ for all $n$ so: $x_n(x_n+\frac{1}{n})<1$ $x_n>0$ so from here we get: $\frac{n-1}{n}<x_n<\frac{\sqrt{4n^2+1}-1}{2n}$ Now we will check a limit when $n\to \infty$ $\lim\limits_{n\to \infty} \frac{\sqrt{4n^2+1}-1}{2n}-\frac{n-1}{n}$ $=\frac{\sqrt{4n^2+1}-2n+1}{2n}$ $=\frac{n(\sqrt{4+\frac{1}{n^2}}-2+\frac{1}{n})}{2n}=0$ So we get that the limit of the diffrence of the limits of $x_n$ is $0$! Hence, there is one possible value for $x_n$ when $n$ is big enough so there is one possible value for $x_1$ when $0<x_n<x_{n+1}<1$ for all $n$ and we are done.

Sorry to bump this but I'm not sure if the previous solution was very rigorous. I mean isn't that the same as doing $ 1-\frac{n-1}{n} = 1/n$ which approaches $ 0$ as $ n$ approaches infinity?.

Indeed, the above presumed proof proves nothing. For some terse solution, see John Scholes' proof at the kalva cache.

I notice that no complete solutions have been given in this topic, so I offer mine below. By recursive substitution, one can write $x_n=P_n(x_1)$ , where $P_n$ is a polynomial with non-negative coefficients and zero constant term. Thus, $P_n(0)=0$, $P_n$ is strictly increasing in $[0,+\infty)$ , and $\displaystyle \lim_{x_1 \rightarrow + \infty} P_n(x_1)=+\infty$. We can therefore define the inverse $P_n^{-1}$ of $P_n$ on $[0,+\infty)$. It follows that $x_1=P_n^{-1}(x_n)$, $P_n^{-1}(0)=0$, $P_n^{-1}$ is strictly increasing in $[0,+\infty)$, and $\displaystyle \lim_{x_1 \rightarrow + \infty} P_n^{-1}(x_1) =+\infty$. Denote by $\displaystyle a_n=P_n^{-1}(1-\frac{1}{n})$ and $b_n=P_n^{-1}(1)$. By the monotonicity of $P_n^{-1}$ we have $a_n<b_n$ for each $n$. Note that: (a) $\displaystyle x_n<x_{n+1} \Leftrightarrow x_n>1-\frac{1}{n} \Leftrightarrow P_n^{-1}(x_n)>P_n^{-1}(1-\frac{1}{n}) \Leftrightarrow x_1>a_n$; (b) $\displaystyle x_n<1 \Leftrightarrow P_n^{-1}(x_n)<P_n^{-1}(1) \Leftrightarrow x_1<b_n$. Thus, $0<x_n<x_{n+1}<1,\forall n$ holds if and only if $a_n<x_1<b_n,\forall n$, or $\displaystyle x_1 \in \bigcap_{n=1}^{+\infty}(a_n,b_n)$. We need to show that $\displaystyle \bigcap_{n=1}^{+\infty}(a_n,b_n)$ is a singleton. We have: (c) if $x_1=a_n$, then $x_n=1-\frac{1}{n}$, which implies that $x_{n+1}=1-\frac{1}{n}<1-\frac{1}{n+1}=P_{n+1}(a_{n+1})$, and $x_1<a_{n+1}$. It follows that $a_n<a_{n+1},\forall n$; and (d) if $x_1=b_n$, then $x_n=1$, which implies that $x_{n+1}=1+\frac{1}{n}>1=P_{n+1}(b_{n+1})$, and $x_1>b_{n+1}$. It follows that $b_n>b_{n+1},\forall n$; and Thus, $a_n<a_{n+1}<b_{n+1}<b_n, \forall n$. Therefore, the two sequences $\{a_n\}_{n=1}^{+\infty}$ and $\{b_n\}_{n=1}^{+\infty}$ converge, and their limits $a$ and $b$ satisfy $a \leq b$. Hence, $\displaystyle \bigcap_{n=1}^{+\infty}(a_n,b_n)=[a,b]$ is non-empty, which demonstrates the existence of $x_1$. Now, suppose that $a \leq x_1 \leq x_1' \leq b$. We have $x_{n+1}'-x_{n+1} = (x_n'-x_n)(x_n'+x_n+\frac{1}{n}) \geq (x_n'-x_n)(2-\frac{1}{n}) \geq (x_n'-x_n)$ for each $n$, so that $x_n'-x_n \geq x_1'-x_1$ for each $n$. However, $1-\frac{1}{n}<x_n \leq x_n'<1$, so that $0 \leq x_n'-x_n<\frac{1}{n}$, which implies that $\displaystyle \lim_{n \rightarrow +\infty}(x_n'-x_n)=0$. Therefore, $x_1' \leq x_1$, proving unicity.

For each $n \ge 1$ let $I_n$ be the interval of real numbers $0 \textless x \textless 1$ such that if $x=x_1$, we will have $x_n \textless x_{n+1} \textless 1$. (That is, the values of $x_1$ which make $x_{n+1}$ "work"). We must prove these intervals intersect at one point. First, I prove they intersect. Notice that as $x_1$ increases, $x_n$ and $x_{n+1}$ do too. So if $I_n=(a_n,b_n)$ it is easy to see $a_n=x_1$ will give $x_n=x_{n+1}=1-1/n$ and $b_n=x_1$ will give $x_{n+1}=1$ (the "extremal" values $x_{n+1}$ can take are given by the extremal values $x_1$ can take). But if $a_n=x_1$ we see that $x_{n+1} \textless 1-1/(n+1)$ and so $x_{n+1} \textgreater x_{n+2}$ and similarly $x_1=b_n$ gives $x_{n+1}=1$ which gives a $x_{n+2}$ too big. So basically when $x_1$ only barely works for $x_{n+1}$, it won't work for $x_{n+2}$. So $I_{n+1}$ is a subset of $I_n$. So we have an infinite sequence of open intervals, each contained inside the previous one. Therefore their intersection must contain at least one number. This guarantees the existence of $x_1$. But, we must prove that $x_1$ can't take two different values. Suppose it could. Suppose $x_1=a,b$ both work. Let $x_i$ be the sequence generated by $a$ and $x_i'$ the sequence generated by $b$, and let $l_i=x_i'-x_i$ (wlog $a \textless b$). Then we see $l_{n+1}=l_n(2x_n+l_n+1/n) \textgreater l_n$ for $n \ge 2$ since $2x_n \textgreater 1$ since $x_n \textgreater 1-1/n \ge 1/2$. So there exists a constant $C$ such that $x_k' \ge x_k+C$ for $k \ge 2$. But since $x_k, x_k' \in (1-1/k, 1)$, this is impossible. So we're done.

Interesting real analysis problem!