Let $n$ be a natural number, and set $f(x)=1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!}$ in $\mathbb{Q}[x]$. Determine the gcd of $f(x)$ and $f'(x)$. Then, determine for which natural numbers $n$ the polynomal $f(x)$ has multiple roots.
I've calculated $f'(x)=1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^{n-1}}{(n-1)!}$. I then tried to do polynomial long division by examining just a few of the highest degree terms and a few of the lowest degree terms, but that didn't seem to be getting me anywhere.
Here $f(x)$ and $f'(x)$ has no common roots:
If $f(u)=f'(u)=0$, then we must have $u^n/n!=0$; that is, $u=0$. Nonetheless, $f(u)=f'(u)=1\neq 0$, a contradiction.
Therefore, $\gcd(f(x),f'(x))=1$.
Note that if the gcd of $f$ and $f'$ has degree higher than $1$, then $f(x)$ and $f'(x)$ must have at least one common root in $\mathbb{C}$.