Prove $f=1+x+x^2+x^3+\cdots+x^n$ has no multiple roots. Prove $f=1+x+x^2+x^3+\cdots+x^n$ has no multiple roots.
My attempt:
Consider the polynomial $g=(x-1)(1+x+x^2+x^3+\cdots+x^n)$
As $f\mid g, g$ all the roots of $f$ are roots of $g$.
This means I have to prove the statement:
If $c$ is a multiple root of $f \Longrightarrow c$ is multiple a root of $g$.
Equivalently:
If $c$ is not a multiple root of $g \Longrightarrow c$ is not a multiple root of $f$.
I'll try to prove this last statement:
Suppose $\exists c$ such that $c$ is a multiple root of $g$, then 
$$g(c)=0 \implies g'(c)=0$$
$$
g=(x-1)(1+x+\cdots+x^n) \implies  g'=nx^n
$$
$$
g'(c)=nc^n=0 \implies c=0
$$
But 
$$
g(0)\neq 0
$$
And we've found the contradiction: $\not \exists c: g(c)=g'(c)=0$. Then $f$ has no multiple roots.
Is this correct? 
Also: Assuming this is correct, is it possible to prove $\sum_{k=0}^n \frac {x^n}{n!} $ has no multiple roots in a similar manner?
 A: All roots of $f(x)$  also are  roots of $(x-1)f(x)= x^{n+1}-1.$ Since    $x^{n+1}-1$ is coprime  with its derivative $(n+1)x^n$  then $x^{n+1}-1$ has  no multiple roots. Thus and $f(x)$ has no multiple roots.
A: Your proof is correct. Here is an alternative.
We have $f(1) = n+1$, so $1$ is not a root, and similarly, $f(0) = 1$, so $0$ is not a root.
Since $f(x) = {1 -x^{n+1} \over 1-x}$, if $x$ is a root of $f$, then $x^{n+1} = 1$, so just need to show that
$f'(x) \neq 0$ whenever $x^{n+1} = 1$.
Since $f'(x) = {1 -x^{n+1} \over (1-x)^2} - (n+1) {x^n \over 1-x }$, we see that
$f'(x) = - (n+1) {x^n \over 1-x } \neq 0$ whenever $f(x) = 0$.
For your other question, it is not clear what you mean by a similar manner.
Let $f_n(x) = 1+x+ {x^2 \over 2!}+\cdots+{x^n \over n!}$. We note that
$f_n(0) = 1$, so $0$ is not a root. Then we have
$f_n(x) = f_n'(x)+ {x^n \over n!}$, and so if $f_n(x) = 0$, we must have $f_n'(x) \neq 0$, hence any root is simple.
A: If you are just talking about real roots, it depends on the value of n:
If n is odd, the root is -1 but if n is even, no real roots exist.
However if you are talking about all possible roots, then there exist exactly n roots, no matter the value of n. The roots of these particular equation are commonly referred to as the nth roots of unity
