KVPY-2015

Let the polynomial be $P(x)=a_nx^n+a_{n-1}x^{n-1}\cdot\cdot\cdot+a_2x^2+a_1x+a_0$ So $P\prime(x)=na_nx^{n-1}+\cdots+3a_3x^2+2a_2x+a_1$ $P(x)-P\prime(x) \text{ has only one term containing } x^n \text{ and its coefficient is } a_n \implies a_n=1.$ Since the coefficients of all other powers of $x$ are $0$ , we get this recursive sequence $a_{n-1}=n a_n,$ Solving for it, $\large{\dfrac {a_n}{a_0}=\dfrac{1}{n!}} \implies a_0=n!$
Now, $P(0)=a_0= {\boxed{n!} }$

$p'(x) - p(x) = -x^n$

The above Differential Equation is of form $f'(x) + Pf(x) = q(x)$

The Integrating Factor is given by $e^{\int Pdx} = e^{\int (-1)dx} = e^{-x}$

Multiplying the DE by Integrating Factor, we get

$e^{-x}p'(x) - e^{-x}p(x) = -e^{-x}x^n$

$\dfrac {d}{dx} ( e^{-x}p(x) ) = -e^{-x}x^n$

Integrating on both sides

$\displaystyle \int \dfrac {d}{dx}(e^{-x}p(x))dx = -\displaystyle \int e^{-x}x^n dx$

$p(x)e^{-x} =- \displaystyle \int e^{-x}x^n dx$

Clearly at $x = \infty$ we get $p(\infty) = 0$

Putting limits from $0$ to $\infty$ for above integral,

$p(\infty)e^{-\infty} - p(x)e^{-x} = -\displaystyle \int_0^\infty e^{-x}x^n dx$

The above integral is Gamma Function $\Gamma (•)$ where $\Gamma (m) = \displaystyle \int _0^\infty t^{m-1} e^{-t} dt$

Using this we get,

$p(x) = \dfrac {\Gamma (n+1) = n! }{e^x}$

At $x = 0$ , $p(0) = \dfrac {n! }{e^0} = \boxed{n!}$