$n_1n_2...n_{11}-1$ is a product of two consecutive natural numbers. 
Prove that there exist distinct, pairwise coprime natural numbers $n_1,n_2,...,n_{11}$ such that $n_1n_2...n_{11}-1$ is a product of two consecutive natural numbers.

Let $N = n_1n_2...n_{11}$ My approach was to firstly notice that none of the $n_1n_2...n_{11}$ is an even integer, otherwise $n_1n_2...n_{11}-1$ would be odd. Then I thought about using $CRT$ \begin{align} N&=1\mod x, \\ N&=1 \mod x+1,\end{align} $x,x+1$ to conssecutive natural numbers. By $CRT$ we know that this system of congruences has a solution. Now I want to prove that the sol. is $0$ $\mod x(x+1)$ and deduce that $n_1n_2...n_{11}-1 = x(x+1)$ is true. How to proceed? Any help appreciated.
 A: I assume you want to avoid the trivial solutions where some of the $n_i$ are $=1$.
Claim. Let $S$ be the set of all primes $p$ for which  $x^2+x+1\equiv 0\pmod{p}$ has a solution. Then $S$ is infinite.
Proof. Assume otherwise. Then  $S=\{p_1,\ldots, p_n\}$ is a finite set of primes. Let $y=p_1\cdots p_n$ and $z=y(y+1)+1$. Then  $z\ge 1\cdot 2+1=3>1$. Let $p$ be a prime divisor of $z$. Then $p\in S$ as witnessed by $x=y$. But then $z\equiv 1\pmod p$, contradicting $p\mid z$. $\square$
Pick $11$ primes $p_1,\ldots , p_{11}\in S$. For each of them, find $x_i$ with $x_i(x_i+1)\equiv -1\pmod {p_i}$ (which is possible because $p_i\in S$). Use CRT to find a natural number $x$ such that $x\equiv x_i\pmod {p_i}$.
Then $x(x+1)\equiv x_i(x_i+1)\equiv -1\pmod{p_i}$ for $1\le i\le 11$. So the number $N:=x(x+1)+1$ has at least $11$ different prime divisors. Thus $N$ can clearly be factored into $11$ pairwise co-prime numbers $>1$ (which are a fortiori pairwise distinct). Of course, $11$ plays no special role and can be replaced by any natural number.

Remark: In my first write-up, I had a wrong expression in $x$ instead of $x^3+x+1$. With the correct expression, $x^2+x+1$, it is of course clear from $x^3-1=(x-1)(x^2+x+1)$ that $x^2+x+1\equiv 0\pmod p$ implies $x^3\equiv 1\pmod p$, and conversely any non-trivial solution of the latter is a solution of the former. From this we read off that $S$ consists precisely of $3$ and the primes $p\equiv 1\pmod 3$. (So essentially the proof of the claim above turns into a proof of the fact that there are infinitely many such primes). It comes to no surprise that the solution by @Oleg567 involves the first such primes (only $1117$ is larger).
A: The following is an answer to the old version of the question, where $n_{i}$ are not necessarily distinct:
As lulu comments, you may take the the example $n_{1},...,n_{10} = 1$ and $n_{11} = 7$.
Note that in the same fashion, it is possible to prove in general that there are infinitely many.
Take $M=k(k+1)+1$, for any $k$, with $n_{1},...n_{m-1}=1$, and $n_{m}=M$.
A: One of (not too large) examples:
$$
n= 303750157,
$$
$$
n(n+1) = 3 \times 7 \times 13 \times 19 \times 31 \times 37 \times 43 \times 61 \times 67 \times 79 \times 1117 -1,
$$
where all the factors in RHS are prime numbers.
Another examples:

$n = 540178291$,

$n = 876805855$,

$n = 956495446$,

$\ldots$
