The solution of the polynomial $(x−1)(x−2)⋯(x−n)−1$ is irreducible in $\mathbb{Z}\left [ x \right ]$ for all $n≥1$ is in here.
I think it's no problem that do the same thing on $\mathbb{Q\left [ x \right ] }$ and $\mathbb{R\left [ x \right ]}$ and get the same result that the polynomial $(x−1)(x−2)⋯(x−n)−1$ is irreducible.
But my question is on $\mathbb{R\left [ x \right ]}$, the polynomial is continuous in $\mathbb{R}$, let $$f(x):=(x−1)(x−2)⋯(x−n)−1$$ Obviously,$f(1)=-1<0;\;f(n+1)=n!-1>0$, so there is a solution between $0$ and $(n+1)$ which means the polynomial is reducible. There's a contradiction, so which part is wrong. Thank you!
The logic doesn't generalize even to $\mathbb{Q}[x]$.
What if $f(k) = \frac{1}{g(k)}$? This would be entirely permissible in $\mathbb{Q}$ since $k$ is rational and hence $g(k)$ is too, if $g \in \mathbb{Q}[x]$.
Of course, $1/n$ is a non-integer for integers $n$ (except $\pm 1$). Since $g \in \mathbb{Z}[x] \implies g(k) \in \mathbb{Z}$ (or more broadly, the evaluation of a polynomial over a ring, at an element in that ring, gives an element in that ring) this where they draw their conclusion in the linked post: it is clear the product must evaluate to $-1$, and the factors must be integers since they are polynomials over $\mathbb{Z}$ too.
Of course, then, this need not hold for polynomials over $\mathbb{Q}$.
Some playing around in WolframAlpha does suggest irreducibility over $\mathbb{Q}$ regardless for $n \ge 2$, where nontrivial cases are concerned. My instinct is suggesting perhaps the rational root theorem may help in proving that irreducibility holds (i.e. prove that $n!-1$ is not a root for any $n$).
It does not hold over $\mathbb{R}[x]$, as you have seen.