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?

Let $f(x)= 1+x+\frac{x^2}{2!}+\ldots +\frac{x^6}{6!}$
Differentiating $f(x)$:
$$f'(x)= 1+x+\frac{x^2}{2!}+\ldots+\frac{x^5}{5!}$$
Consider $a$ as one real root for $f'(x)=0$, i.e. $f'(a)=0$. Obviously $a\ne 0$.
Now, $f(a)= 0+ \frac{a^6}{6!}>0$. This means that all the potential local minima of $f(x)$ are already positive. Hence, there are no real roots for $f(x)=0$.
