How surprising is this result on the diophantine equation $x(x+1)...(x+n-1) = y^n+k$? A number of years ago, I proved the following result:
For any integer $k$, the number of positive integral solutions to
 $x(x+1)...(x+n-1) = y^n+k$
with $n \ge 3$ is finite 
(i.e., there are only a finite number of $(x, y, n)$ satisfying this equation
for any $k$).
It is pretty clear that for any fixed $k$ and $n$ there are only a finite number of $x$ and $y$ (you can prove that $y \le |k|$), but the fact that there are only a finite number of $n$ came as a surprise to me. I initially proved that $n < e|k|$ and later derived much stricter bounds.
The way I phrased this is "The product of $n$ consecutive integers is almost never close to an $n$-th power."
My question is whether this result is surprising?
Thanks.
 A: The result is not surprising in the sense that standard probabilistic heuristics suggest it to be true. In the interval $[2^d, 2^{d+1})$ there are $2^d$ numbers. Of those numbers, $O(d \cdot 2^{d/3})$ are perfect third powers or higher, so the probability that a number in this interval is a perfect third power or higher is $O(d \cdot 2^{-2d/3})$. The probability that a number in this interval is of the form $x(x+1)...(x+n-1) - k$ for some $n \ge 3$ should be approximately the same, and if we assume that the two events are independent, then the probability that a solution exists to your Diophantine equation with $y^n \in [2^d, 2^{d+1})$ is $O(d^2 \cdot 2^{-4d/3})$.
The sum of these probabilities over all $d$ converges. A heuristic application of Borel-Cantelli then suggests that the number of solutions is finite. 
Of course, this argument also applies, incorrectly, to $n = 2$. Here there is algebraic structure which violates the independence assumption. But this structure doesn't generalize to higher $n$, and I guess this can be made precise by Faltings' theorem for fixed $n$. 
Speaking of which, Scott Carnahan's answer to this MO question about the Mordell conjecture may be more convincing to you, as it correctly excludes the case $n = 2$. (Take the sum over all $d \ge 3$, in the notation of his answer.) 
A: BTW, here is how I converted the bound on $y$ to a bound on $n$:
Since $2(n/e)^n < n!$,
$$2(n/e)^n < n! \le  x(x+1)...(x+n-1) = y^n+k.$$
Using the corrected bound on $y$ of $y \le |k|$,
$$2(n/e)^n < |k|^n + |k| \le 2|k|^n$$
so $n/e < |k|$ or $n < e|k|$.
I did this over 40 years ago, but I still remember thinking that this was magic,
somehow pulling the result out of thin air.
A: It is a result of Erdos and Selfridge that the product of consecutive integers is never a power (with trivial exceptions). Perhaps something in their paper is of use here. 
