Perturbing integer roots The polynomial
$$(x-a_1)(x-a_2)\cdots(x-a_n) - x^{n-1} = 0$$
for arbitrary integers $a_i$ has come up in a project.  Is the root with the greatest modulus always real?  Evidence from several choices of the $a_i$ for small $n$ suggests that the answer is yes...
I titled this perturbing the roots since, without the $x^{n-1}$ term, the zeros would simply be the $a_i$. This feels like the sort of thing that was better known when "theory of equations" classes were a mainstay of mathematics training.

Edit: As per Chris Sanders below, the $a_i$ should be positive integers.
 A: $(x+1)^2-x$ has no real roots.
$(x+1)^3-x^2$ has exactly one real root with modulus smaller than the modulus of either of the other roots.
https://www.wolframalpha.com/input?i=%28x%2B1%29%5E3-x%5E2
EDIT:
I think you might want to require that all $a_i$ be positive integers.
In that case, have a look at my other answer.
A: Let $p(z)=(z-a_1)\ldots(z-a_n)$,
where $a_1,...,a_n$ are all positive integers.
Without loss of generality, $a_1\leq a_2\leq\ldots\leq a_n$.
$\\$
Denote by $r$ the largest real root of $p(x)-x^{n-1}$.
Now, because $p$ is of degree $n$, you get $\displaystyle\lim_{x\to\infty}p(x)-x^{n-1}=\infty$,
but $p(a_n)-a_n^{n-1}=-a_n^{n-1}<0$.
Therefore, there must be a real root of $p(x)-x^{n-1}$ to the right of $a_n$. Hence, $r>a_n$.
Let's compare $p(rw)$ and $(rw)^{n-1}$, where $w$ is on the unit complex circle $S^1$; in other words $|w|=1$.
Now $|(rw)^{n-1}|=r^{n-1}$, but what about $p(rw)=(rw-a_1)\ldots(rw-a_n)$?
For $w\in S^1\setminus\{1\}$, you find that $|rw-a_j|>r-a_j$. This follows from the "equality conditions" of the triangle inequality. So actually, $|p(rw)|=|rw-a_1|\ldots|rw-a_n|>(r-a_1)\ldots(r-a_n)=r^{n-1}$.
This means that of the roots of $p(z)-z^{n-1}$, there is only one of the form $rw$ for $w\in S^1$, and that root is $r$ itself.
$\\$
Let $t$ be a real number greater than $r$.
Then, because $r$ is the largest real root of $p(x)-x^{n-1}$,
it follows that $p(t)>t^{n-1}$.
Again, for $w\in S^1$, you get
$|p(wt)|=|wt-a_1|\ldots|wt-a_n|\geq(t-a_1)\ldots(t-a_n)>t^{n-1}=|(tw)^{n-1}|$.
We conclude that $p(z)-z^{n-1}$ has no complex roots with modulus greater than $r$.
