Prove that $f(x)=0$ has no repeated roots

$$\text{If } f(x)=\frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}+\cdots+x+ 1\text{, then show that } f(x) = 0\\ \text{ has no repeated roots.}$$

I tried solving this question and I think I have come up with a proper answer. I need some verification.

My solution/attempt

First, we need to prove a theorem.

Theorem 1: If a polynomial function $f(x)$ has a repeated root (say $a$, i.e. $f(a)=0$), then $f'(a)=0$.

Proof: We assume that $f(x)$ has a degree of $n \geq 2$. Since $a$ is a factor of $f(x)$, we can write:

$$f(x)=(x-a)^m\cdot h(x) \tag{1}$$ where $2 \leq m \leq n$ and $h(x)$ is a polynomial of degree $n - m$

On differentiating $(1)$, we can write $$f'(x) = m(x-a)^{m-1}\cdot h(x) + (x-a)^m\cdot h'(x)\tag{2}$$

Plugging in $x=a$, we obtain $$f'(a)=0+0=0$$

Thus, theorem 1 is true.

Now, from the question, we have $$f(x)=\frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}+\cdots+x+ 1\tag{3}$$ If we plug in $x=0$, we get $f(0)=1 \neq 0$, for any value of $n$. Hence $0$ is not a root of $f(x)$.

On differentiating, we get $$f'(x)=\frac{x^{n-1}}{(n-1)!}+\frac{x^{n-2}}{(n-2)!}+\cdots+x+ 1\tag{4}$$

From $(3)$ and $(4)$, we obtain $$f(x)-f'(x)=\frac{x^n}{n!} \tag{5}$$

Suppose $c \neq 0$ is a root of $f(x)$. Then from $(4)$, we have

$$f'(c)=-\frac{c^n}{n!} \neq 0\tag{6}$$

Hence, from theorem 1, we can conclude that $f(x)=0$ has no repeated roots.

Q.E.D.

My question

I am a beginner in this field. Did I do all the steps correctly? Are there any points I need to take care of?

Let $e_{n}(x)=\sum_{k=n+1}^{\infty} {x^{k}\over{k!}}$, therefore $f_{n}(x)=e^{x}-e_{n}(x)$. To find repeated roots we must solve $f_{n}(x)=f_{n}^{'}(x)=0$ which means $e_{n}(x)=e_{n}^{'}(x)=e^{x}$. Also differentiating $e_{n}(x)$ gives us $e_{n}^{'}(x)=\sum_{k=n+1}^{\infty} k{x^{k-1}\over{k!}}=\sum_{k=n}^{\infty} {x^{k}\over{k!}}=e_{n-1}(x)$. Therefore we have $e_{n}(x)=e_{n-1}(x)$ to solve, which gives us the only answer x=0 which is incorrect when applying it to $f_{n}(x)=0$, and this is what we needed to prove.