Prove that functional equation doesn't have range $\Bbb R.$ Prove that any solution $f: \mathbb{R} \to \mathbb{R}$ of the functional equation
$$ f(x + 1)f(x) + f(x + 1) + 1 = 0 $$
cannot have range $\mathbb{R}$.

I transformed it into 
$$ f(x) = \frac {-1} {f(x + 1)} - 1 = \frac {-1 - f(x + 1)} {f(x + 1)} $$
I tried to evaluate $f(x + 1)$ and $f(x + 2)$ and put them into the transformed equation:
1) after $f(x + 1)$
$$ f(x) = \frac {1} {-1 - f(x + 2)}$$
2) after $f(x + 2)$
$$ f(x) = \frac {1} {\frac {-1}{f(x + 3)}} = -f(x + 3) $$
What am I supposed to do next?

Assume that the range of $f$ is all of $\Bbb R$. Then there are $a,b\in\Bbb R$ with $f(a)=0$ and $f(b)=-1$. But then 
$$ 1=f(a)f(a-1)+f(a)+1=0$$
and 
$$ 1=f(b+1)f(b)+f(b+1)+1=0,$$
both of which are absurd.
