If I can calculate $6!$ as $1 \times 2 \times 3 \times 4 \times 5 \times 6$; why is incorrect calculate the factorial of $6.1$ as $1.1 \times 2.1 \times 3.1 \times 4.1 \times 5.1 \times 6.1$?
Please note that I'm not asking for the right way to calculate $6.1!$
I'm asking why is incorrect calculate the factorial in this way.
$6!=1\times2\times3\times4\times5\times6$ is a product of six numbers, and $7!$ is a product of seven numbers, so logically $(6.1)!$ should be a product of $6.1$ numbers, not a product of six numbers. That's one way to see why it's the wrong way to compute $(6.1)!$.
Defining $(6+r)!$ to be $(1+r)(2+r)\cdots(6+r)$ runs into trouble for $r>1$, e.g., it gives $8!=(6+2)!=(1+2)(2+2)\cdots(6+2)=20160$, which is incorrect.
There are also problems going backwards: should $(5.9)!$ be $.9\times1.9\times\cdots\times5.9$ (since $5.9=6-0.1$), or should it be $1.9\times2.9\times\cdots\times5.9$ (since $5.9=5+0.9$)?