Is there a closed form solution for this polynomial? Let $1\leq j \leq n$, the polynomial is
$$ \prod_{i=1;i \neq j}^{n} (x-i) = \frac{\prod_{i=1}^{n} (x-i) }{x-j} = \frac{(x-1)(x-2) ....(x-n)}{x-j} $$
I've tried looking for it but haven't found anything on this. Wolfram only details this with the Pochammer symbol, but that isn't very useful. No, this is not homework. I would like to know if there was a paper on this that maybe derives the coefficients of this polynomial if it is even possible?
I found that Vieta's formulas for $\prod_{i=1}^{n} (x-i)$ simplify the first few terms, but afterwards it gets very complicated.
$p_n(x)=a_n x^n + a_{n-1} x^{n-1} + ... + a_0$
$a_n=1$
$a_{n-1}=\sum_{i=1}^{n} i =\frac{n(n+1)}{2}$
$a_{n-2}=\sum_{i,j;i<j}^{n} ij=\sum_{i=2}^{n} \frac{(i-1)(n-i-1)(n+i)}{2}=\frac{n(n^3-6n^2+3n+2)}{8}$
...
$a_0=n!$
You're not going to find a closed form for these numbers. I can give you a formula for the coefficient of $x$ as proof: $$(-1)^n\frac{n!}j\left(H_n-\frac1j\right)$$ where $H_n$ is the $n$th harmonic number, for which no closed formula is known.