Repetition can make you go nuts!!

First let us consider $f(x)$ has a repeated root say at $x=a$ . Since it is a repeated root we must have $f(a) = 0$ and $f'(a) = 0$ .

If $f(a) = 0$ ,we must have $1+\frac{a}{1!}+\frac{a^{2}}{2!} + \frac{a^{3}}{3!}+..... \frac{a^{n}}{n!} = 0$ --- (1)

If $f'(a) = 0$ ,we must have $1+\frac{a}{1!}+\frac{a^{2}}{2!} + \frac{a^{3}}{3!}+..... \frac{a^{n-1}}{(n-1)!} = 0$ ---(2)

Subtracting these two equations we obtain $\frac{a^{n}}{n!} = 0$ $a = 0$ .

Substituting $a=0$ in any of the above two equations we obtain $1=0$ ...an obvious contradiction.

Hence $f(x)$ can never have a repeated root.