Usually, I would use Rolle's Theorem. Every polynomial is continuous and differentiable.
Using Rolle's Theorem, derivatives of functions have $n-1$ real roots:
Between every two consecutive roots $a$ and $b$, there will be point $c$, where $a<c<b$, that $f'(c)=0$. If the roots aren't consecutive, then the upper bound will be $n-1$.
But then $f(x)=x^2+1$ has no real roots, but its derivative has one real root: $f'(x)=2x=0, x=0$
Now my question is: How do I prove that there are no complex roots with Rolle's Theorem?
Note: This was in last years Calc exam, and im revising for my next one, so plz don't come at me saying this was in my lectures :)
Notice that you already have that, since $P_n$ is a $n$-degree polynomial, then it has at most $n$ roots (with repetition and including complex numbers). The same is true for $P_{n-1}$: polynomials of degree $n-1$ have $n-1$ roots, once you include repeated and complex roots. (This is known as the fundamental theorem of algebra, or sometimes taken as a consequence of it.)
$P_n$ is given to have $n$ real roots. Using Rolle's theorem, you have deduced that $P_{n-1}$ (i.e. $P_n'$) has $n-1$ real roots. Since it is degree $n-1$, it cannot have any more roots than that.
We are not concerned with this case. We are only concerned with the premise that IF $f$ is a polynomial, of degree $n$, with $n$ real roots, THEN we want to show that $f'$ is of degree $n-1$ with $n-1$ real roots.
$f(x) = x^2+1$ does not fit into the "if" statement, so we ignore that.