Proving that $f_n(x)=\frac{(x+1)^n-x^n-1}{x(x+1)}$ is strictly positive for odd integers $n\geq 5$ 
Let $n\geq 5$ be an odd integer. I'd like to prove that the function $$f_n(x)=\frac{(x+1)^n-x^n-1}{x(x+1)}$$ is strictly positive on $\mathbb{R}$.

(I already figured out that the numerator is divisible by the denominator, so $f_n(x)$ is in fact a polynomial of degree $n-3$.
I tried to find the minimum by differentiation, but I have difficulty finding the location and the value of the minimum. Another thing I know is that $f_n(x)$ is palindromic, but it doesn't seem to give any inspirations. Does anyone have ideas?
Thanks in advance!

I will treat $f$ as  polynomial of degree $n-3$.
If $f$ takes both positive and negative values then it must vanish at some point. But $(x+1)^{n}=x^{n}+1$ has no  solution in the real line other than $-1$ and $0$: Its derivative is strictly positive in $(-\frac 1 2 , \infty)$ and strictly negative in $(-\infty, -\frac 1  2 )$. So it attains its minimum at $-\frac  1 2$ and there are exactly two values at which it vanishes. Of course, these are $0$ and $-1$. Now $f_n(0)\neq 0$ since $x^{2}$ is not  factor of the numerator and $f_n(-1)\neq 0$ since $(1+x)^{2}$ is not  factor of the numerator. Hence, $f$ does not change sign. Also, $f(1)>0$ so $f(x) >0$ for all $x$.
