Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$. 
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.

Attempts so far:
Used Descartes signs stuff so possible number of real roots is $6,4,2,0$
tried differentiating the equation $4$ times and got an equation with no roots hence proving that above polynomial has $4$ real roots.
But using online calculators I get zero real roots. Where am I wrong?

First, rescale by $720$ to get integer coefficients: $$x^6+6x^5+30x^4+120x^3+360x^2+720x+720$$
Now repeated completion of binomial powers:
$$\begin{align}
&\phantom{{}={}}(x+1)^6+15x^4+100x^3+345x^2+714x+719\\
&=(x+1)^6+15(x+5/3)^4+95x^2+\frac{3926}{9}x+\frac{16288}{27}\\
\end{align}$$
You could complete the square again on this last quadratic and you will be left with a positive constant, or you can just compute its discriminant to see that the quadratic itself has no roots (and has a positive quadratic term, so is therefore positive). So $$(x+1)^6+(x+5/3)^4+q(x)$$ is positive.
