Real roots of $ 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots +\frac{x^n}{n!} $ 
Let $Q_n(x)$ be the degree $n$ polynomial
  $$ 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots +\frac{x^n}{n!} $$
  How many real roots does the equation $Q_n(x)=0$ have?


My attempt:
It is obvious that $Q_n(x)$ will have all its real roots in the negative part of the real line if there is any. Also, we notice that if $n$ is odd, then there is at least one real root by the complex conjugate root theorem. So I conjecture that there is exactly one root for $n$ odd and there is no root for $n$ even. 
However, I don't know how to analyze $Q_n(x)$. All I can do is to take derivative and this does not provide more useful information. Any hint is appreciated! Thanks.

Your conjecture is correct and it can be proved by induction.
The statement is trivially true if $n=1$. Assume that it is true for a certain $n$. If $n$ is even, then $(\forall x\in\mathbb R):Q_n(x)>0$. So, $Q_{n+1}$ is strictly increasing (note that $Q_{n+1}'(x)=Q_n(x)$) and therefore has at most one real root. But every polynomial whose degree is odd has at least one root. So, it has exactly one root.
And if $n$ is odd, then $Q_{n+1}$ first decreases and then increases. So, it has an absolute minimum, which is attained at the point $x_0$ such that $Q_n(x_0)=0$. But\begin{align}Q_{n+1}(x_0)&=Q_n(x_0)+\frac{{x_0}^{n+1}}{(n+1)!}\\&=\frac{{x_0}^{n+1}}{(n+1)!}\\&>0,\end{align}since $n+1$ is even and $x_0\neq0$ (since $Q_n(0)=1\neq0$).
