If $P$ is a monic polynomial of degree $2015$ such that $P(i)=2014-i\quad\forall i=0,...,2014$ & $P(2015)=n!-a$, find $n+a$

Question

If $P(x)$ be a polynomial of degree $2015$ with leading coefficient $1$ such that $P(0)=2014, P(1)=2013, P(2)=2012,...,P(2014)=0$ and $P(2015)=n!-a$, where $n$ and $a$ are natural numbers, find $(n+a)$.

Attempt:

Obviously the constant is $2014$. The sum of the coefficients is $2013$. And it can be observed that $P(x+1)-P(x)=1$

But then what?

Answer 1

We define a new function $g(x) = p(x) + x - 2014$.

Now, we know degree of $p(x)$ is 2015. Hence, degree of $g(x)$ is also 2015

Now, $g(x)$ has roots $0,1,2,3...2014$. Hence, we write $g(x)$ as

$$g(x) = x(x-1)(x-2)....(x-2014)$$ $$\implies p(x) +x - 2014 = x(x-1)(x-2)...(x-2014) $$ $$\implies p(x) = x(x-1)(x-2)....(x-2014) -x + 2014$$

Hence, $p(2015) = 2015! - 2015 + 2014 = 2015! -1 \implies n+a = 2016$