Consider the following family of polynomials
$$P_n(X)=\sum_{i=0}^n(n+1-i)X^i,\,n\ge 1$$
Let’s write down the first few
$$
\begin{align}
&P_1(X)=X+2\\
&P_2(X)=X^2+2X+3\\
&P_3(X)=X^3+2X^2+3X+4\\
&P_4(X)=X^4+2X^3+3X^2+4X+5
\end{align}
$$
My claim is that this family is a family of irreducible polynomials in $\Bbb{Z}[X]$.
I proved it for $n \le 5$ by the Eisenstein criterion after the change of variable $X=t+1$
For $n=6$ after the change of variable the polynomial writes as follows
$$Q_6(t)=P_6(X-1)=t^6+8t^5+28t^4+56t^3+70t^2+56t+28$$
and the Eisenstein criterion (only workable $p=2$) doesn’t work anymore.
By reducing $\bmod 7$ we prove the claim for $n=6$.
I tested with Mathematica and irreducibility is checked for $n\le 150$.
I noticed
$$P_n(X)=XP_{n-1}(X)+n+1$$
But I am struggling to find a generic proof. I have given up on counterexample. Thanks for your help.
Best Answer
This is not a complete answer, but summarizes three provable cases. For simplicity (and better familiarity with existing posts), let $f_n(x)=x^{n-1}+2x^{n-2}+\dots+(n-1)x+n$.
Case 1: $n+1$ is a prime. As you have noticed, in many cases we can use Eisenstein criterion for $f_n(x+1)$. We can show that this will work when $n+1$ is a prime. We have $$ f_n(x+1)=\sum_{i=0}^{n-1}(n-i)(x+1)^i=\sum_{i=0}^{n-1}(n-i)\sum_{j=0}^{i}\binom{i}{j}x^j=\sum_{j=0}^{n-1}\sum_{i=j}^{n-1}(n-i)\binom{i}{j}x^j $$ and the coefficient at $x^j$ is $$[x^j]f_n(x+1)=\sum_{i=j}^{n-1}(n-i)\binom{i}{j}=\sum_{i=0}^{n-1}(n-i)\binom{i}{j}=\binom{n+1}{j+2}$$
Now having $p=n+1$ a prime, we have $p \mid \binom{n+1}{j+2}$ for $j+2<n+1$, i.e. $j<n-1$. Also for $j=0$ the coefficient becomes $\binom{n+1}{2}=\frac{n}{2}(n+1)$ and it is clearly divisible by $p$ but not by $p^2$, so the polynomial satisfies conditions of Eisenstein criterion for $p$ and is therefore irreducible.
Case 2: $n$ is a prime. The irreducibility can be also shown for $n$ being a prime as discussed in comments. The idea is that all complex roots of the $f_n(x)$ lie outside of the unit circle in a complex plane, which can be shown by looking at roots of $(x-1)f_n(x)=x^n+x^{n-1}+\dots+x-n$. Then having $f_n(x)=p(x)q(x)$ with $n$ prime implies that without loss of generality we can take $p(x)$ such that $|p(0)|=1$. But $|p(0)|=\prod |x_k|>1$ gives a contradiction. More details on this can be found here Explain proof of irreducibility of $x^{p-1} + 2x^{p-2}+ \dots +(p-1)x + p$.
Case 3: $n$ is a prime power. The previous case argument can be modified to work on any prime power. Say we have $n=p^k$, and $f_n(x)=g(x)h(x)$ where coefficients of $g(x),h(x)$ are $a_i$,$b_i$ respectively. For constant coeficient we have $n=p^k=a_0b_0$, so $a_0$ and $b_0$ are non-negative powers of $p$. Then for linear coefficient of $f_n(x)$ we have $n-1=a_1b_0+a_0b_1$. Since $(n-1,n)=1$, one of $a_0,b_0$ must be equal $\pm 1$, say $a_0$. Finally, again as in previous case we reach contradiction by looking at $|g(0)|=\prod |x_k|>1$.
Remaining cases are problematic and I could not find anything about this in literature or prove it. I've tried to get more by looking at the roots but it did not get me much, here is at least an observation I stumbled upon - all complex roots of the polynomial lie in annulus $1<|x_k|<\sqrt[n]{2n+1}$. First inequality is already proven above. For the second inequality notice that $(x-1)^2f_n(x)=x^{n+1}-(n+1)x+n$, and so for a root $z$ we have $$n=|z^{n+1}-(n+1)z|=|z||z^n-(n+1)|.$$ Now being interested only in roots $|z|>1$, this gives $n>|z^n-(n+1)|$, which geometrically means that $z^n$ distance from $n+1$ is less than $n$, so especially $|z^n|<n+1+n$, from which we have $|z|<\sqrt[n]{2n+1}$.
Hopefully someone in the future will have an idea on the remaining cases, but notice that this problem was posted in Putnam and there it was only for $n$ being prime, so this seems like a hard problem.
Update: Literature reference
Okay I have managed to find a reference related to the problem, specifically see Classes of polynomials having only one non-cyclotomic irreducible factor by A. Borisov, M. Filaseta, T. Y. Lam and O. Trifonov (http://matwbn.icm.edu.pl/ksiazki/aa/aa90/aa9023.pdf). It states a conjecture about $f(x)=1+x+x^2+\dots+x^n$, then $f'(x)=1+2x+\dots+nx^{n-1}$ being irreducible for all $n \geq 2$, but that is just a reciprocal of $f_n(x)$, so it is equivalent to this problem. It states that the irreducibility has been proven previously for $n=p-1$ or $n=p^r$, and furthermore also when $n+1$ is squarefree or $n=2p-1$. Then the paper itself claims additional results, specifically: