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\begin{questionparts}
\item $\,$ \begin{align*}
&& f'_n(x) &= 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \cdots + \frac{nx^{n-1}}{n!} \\
&&&= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{n-1}}{(n-1)!} \\
&&&= f_{n-1}(x)
\end{align*}

\item Claim: $f_n(x) > 0$ for all $x > 0$

Proof: (By induction)
Base case: ($n = 1$) $f_1(x) = 1 + x > 1$ therefore $f_1(x) > 0$

Suppose it's true for $n = k$, then consider $f_{k+1}$, if we differentiate it, we find it is increasing on $(0, \infty)$ by our inductive hypothesis. But then $f_{k+1}(0) = 1 > 0$. Therefore $f_{k+1}(x) > 0$ as well.

Therefore by the principle of mathematical induction we are done.

Since $f_n(x) > 0$ for non-negative $x$, if $a$ is a root it must be negative.

\item Suppose $f_n(a) = f_n(b) = 0$ then $f'_n(a) = -\frac{a^n}{n!}$ and $f'_n(b) = -\frac{b^n}{n!}$, but then $f_n'(a) f_n'(b) = \frac{(-a)^n(-b)^n}{(n!)^2} > 0$ since $a < 0, b < 0$.

$_n'(a) f_n'(b)$ is positive, the two gradients must have the same sign (and not be zero). Therefore if they are both increasing, at some point the curve must cross the axis in between. Therefore there is some root $c$ between $a$ and $b$. But then there is also a root between $c$ and $a$ and $c$ and $b$, and very quickly we find more than $n$ roots which is not possivel. Therefore there must be at most $1$ root.

If $n$ is odd there must be exactly one root, since $f_n$ changes sign as $x \to -\infty$ vs $x = 0$.

If $n$ is even then there can't be any roots, since if it crossed the $x$-axis there would be two roots (not possible) and it cannot touch the axis, since $f'_n(a) \neq 0$ unless $a = 0$, and we know $a < 0$
\end{questionparts}

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