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?

Previously, I posted an answer using Sturm Chains. That answer did not give any idea whether this was true for all even-ordered truncations of the Taylor Series for $e^x$.  Here is an answer that works for all even-ordered truncations of the Taylor Series for $e^x$.

Polynomial Product Solution
Define
$$
e_n(x)=\sum_{k=0}^n\frac{x^k}{k!}\tag1
$$
Then
$$
\begin{align}
e_n(x)\,e_n(-x)
&=\sum_{k=0}^n\frac{x^k}{k!}\sum_{j=0}^n(-1)^j\frac{x^j}{j!}\tag2\\
&=\sum_{k=0}^{2n}\sum_{j=\max(0,k-n)}^{\min(k,n)}(-1)^j\binom{k}{j}\frac{x^k}{k!}\tag3\\
&=\sum_{k=0}^{2n}\sum_{j=k-n}^n(-1)^j\binom{k}{j}\frac{x^k}{k!}\tag4\\
&=\sum_{k=0}^{2n}\sum_{j=0}^k\left[(-1)^n\binom{-1}{n-j}-(-1)^{k-n-1}\binom{-1}{k-n-1-j}\right]\binom{k}{j}\frac{x^k}{k!}\tag5\\
&=\sum_{k=0}^{2n}\left[(-1)^n\binom{k-1}{n}+(-1)^{k-n}\binom{k-1}{k-n-1}\right]\frac{x^k}{k!}\tag6\\
&=1+(-1)^n\sum_{k=1}^{2n}\left(1+(-1)^k\right)\binom{k-1}{n}\frac{x^k}{k!}\tag7\\[3pt]
&=1+(-1)^n\sum_{k=1}^n2\binom{2k-1}{n}\frac{x^{2k}}{(2k)!}\tag8
\end{align}
$$
Explanation:
$(2)$: apply $(1)$
$(3)$: swap order of summation, substitute $k\mapsto k-j$,
$\phantom{\text{(3):}}$ swap order of summation again
$(4)$: when $k\lt j\le n$, $\binom{k}{j}=0$, so set the upper limit to $n$
$\phantom{\text{(4):}}$ when $k-n\le j\lt 0$, $\binom{k}{j}=0$, so set the lower limit to $k-n$
$(5)$: $(-1)^n\binom{-1}{n-j}-(-1)^{k-n-1}\binom{-1}{k-n-1-j}=(-1)^j[k-n\le j\le n]$
$\phantom{\text{(5):}}$ and $\binom{k}{j}$ set the limits; we only need to sum $0\le j\le k$
$(6)$: Vandermonde's Identity
$(7)$: the $k=0$ term is $1$ and for $k\ge1$, $\binom{k-1}{k-n-1}=\binom{k-1}{n}$
$(8)$: terms for odd $k$ are $0$; substitute $k\mapsto2k$
For even $n$, $e_n(x)\,e_n(-x)\ge1$. Since $e_n(x)\ge1$ for $x\ge0$, $(8)$ says that
$$
\bbox[5px,border:2px solid #C0A000]{e_n(x)\gt0\quad\text{for even $n$ and all $x\in\mathbb{R}$}}\tag9
$$


Here is a solution that extends the answer by abel.

Differential Equation Solution
Define $e_n(x)$ as in $(1)$. Then
$$
e_n(t)-e_n(t)'=\frac{t^n}{n!}\tag{10}
$$
Solving via integrating factor:
$$
\begin{align}
e^x\int_x^\infty\frac{t^n}{n!}e^{-t}\,\mathrm{d}t
&=e^x\int_x^\infty\left(e_n(t)-e_n(t)'\right)e^{-t}\,\mathrm{d}t\tag{11}\\
&=e^x\int_x^\infty\left(-e_n(t)e^{-t}\right)'\,\mathrm{d}t\tag{12}\\[3pt]
&=e^x\left[-e_n(t)e^{-t}\right]_x^\infty\tag{13}\\[9pt]
&=e_n(x)\tag{14}
\end{align}
$$
For even $n$, it is easy to see that the left hand side of $(11)$ is positive for all $x$, so $(14)$ is also positive. That is,
$$
\bbox[5px,border:2px solid #C0A000]{e_n(x)\gt0\quad\text{for even $n$ and all $x\in\mathbb{R}$}}\tag{15}
$$
