Use Rolle's theorem to find the condition for the polynomial equation $f\left(x\right)=0$ to have a repeated real roots. Hence, find that the equation; $1\frac{x}{1!}\frac{x^2}{2!}\dots \frac{x^n}{n!}=0$, has

By Rolle's theorem, we can say that between any two roots of a polynomial there is always a root of its derivative.

 Thus, if $\alpha$ is a repeated root of a polynomial $f\left(x\right)$, then there must be a root of $f^{\text{'}}\left(x\right)$ in the interval.

$\Rightarrow f^{\text{'}}\left(\alpha \right)=0$

i.e. $f\left(\alpha \right)=f^{\text{'}}\left(\alpha \right)=0$, for $\alpha$ to be a repeated root.

Let $\varphi \left(x\right)=1\frac{x}{1!}\frac{x^2}{2!}\dots \frac{x^n}{n!}$ has a repeated root $\alpha$.

$$\begin{gathered} \varphi \left(\alpha \right)=0 \\ \Rightarrow \varphi \text{'}\left(\alpha \right)=0 \end{gathered}$$

$\Rightarrow 1+\frac{\alpha }{1!}+\frac{{\alpha }^2}{2!}+\dots +\frac{{\alpha }^n}{n!}=0$

and $1+\alpha +\frac{{\alpha }^2}{2!}+\dots +\frac{{\alpha }^{n-1}}{(n-1)!}=0$

Solving above equations, we get

$\frac{{\alpha }^n}{n!}=0$

or $\alpha =0$, thus 0 is the repeated root of $\varphi \left(x\right)=0$.

But, 0 doesn't satisfy $\varphi \left(x\right)$.

$\therefore$ There is no repeated root of $\varphi \left(x\right)=0$.

Hence, the correct option is $1.$