Here is an equation:
$$ e^x\frac{d^n(x^ne^{-x})}{dx^n}=0 $$
Now I want to prove that this equation has $n$ different roots.
I tried to convert the equation to this form:
$$ \sum\limits_{k=0}^{n}C_n^k(-1)^{n-k}\frac{n!}{k!}x^{n-k}=0 $$
But I don't know how to prove that it has exactly $n$ different roots. So what should I do?
A hint is enough!
Rolle's theorem is enough. $f(x)=x^n e^{-x}$ has a root with multiplicity $n$ at the origin and it vanishes at $+\infty$.
$f'(x)$ has a root with multiplicity $n-1$ at the origin, a simple root at some $x_0\in\mathbb{R}^+$ (local maximum for $f$) and it vanishes at $+\infty$. $f''(x)$ has a root with multiplicity $n-2$ at the origin, a simple root somewhere between $0$ and $x_0$ and a simple root somewhere after $x_0$. Use induction and you are done.
The same argument also applies to (shifted) Legendre polynomials, essentially given by $\frac{d^n}{dx^n}\left(x^n(1-x)^n\right)$.
The largest root of a Laguerre polynomial can be estimated using the techniques shown here for Hermite polynomials.