Using induction or otherwise, show that the polynomial
$$P_{n}(x)=1+x+\frac{x^{2}}{2}+\cdots+\frac{x^{n}}{n!}$$
has exactly one real zero if $n$ is odd and none if $n$ is even.
I believe the key here is that $P_{n}'(x)=P_{n-1}(x)$, but I cannot see how to apply this to the problem.
On
For odd $n$: The proposition is true for $n\leq 2$ by elementary algebra. Suppose $n$ is odd and $n\geq 3$ and that the proposition is true for all $m<n.$ Then $P'_n=P_{n-1}$ is never $0,$ and $P_{n-1}(0)=1>0.$ So $P'_n (=P_{n-1})$ is always positive. So $P_n$ is strictly increasing. A monotonic real polynomial of odd degree is a bijection.
I haven't yet had time for the case of even $n.$
We proceed by induction, the case $n=0$ being trivial.
If $n$ is odd: We have $P_n(0)=1$, and $P_n(-n-1)<0$: consider the terms $\frac{x^k}{k!}+\frac{x^{k+1}}{(k+1)!}$ where $k$ is even. Evaluating at $-n-1$ we have $\frac{(n+1)^k}{k!} - \frac{(n+1)^{k+1}}{(k+1)!}$, which has the same sign as $(k+1)(n+1)^k - (n+1)^{k+1}$ which again has the same sign as $k+1 - (n+1)$. Since $k+1\leq n$, we obtain that this is negative. Hence, there is a root somewhere in the interval $[-n-1,0]$ by continuity of $P_n$. Now the derivative of $P_n$ is $P_{n-1}$ so by induction hypothesis $P_{n-1}$ has no root, which means that $P_n$ is monotonic and has thus at most one root, which proves the statement. If $n$ is even: By the induction hypothesis, $P_{n-1}$ has exactly one root, and we know that this root is in $[-(n-1)-1,0]$. We see that $P_n$ tends to $+\infty$ at both $+\infty$ and $-\infty$. Its unique minimum must then be in $[-n,0]$. We can check that $P_n(x)$ is positive for $x\in [-n,0]$ by considering this time the terms $\frac{x^k}{k!}+\frac{x^{k+1}}{(k+1)!}$ where $k$ is odd, and this proves that $P_n$ has no root.