TAOCP 1.2.5 Exercise 25
No, factorial powers do not satisfy a law analogous to the ordinary law of exponents.
Section 1.2.5: Permutations and Factorials
Exercise 25. [**] [M20] Do factorial powers satisfy a law analogous to the ordinary law of exponents, $x^{m+n}=x^m x^n$?
Verified: yes
Solve time: 39s
No, factorial powers do not satisfy a law analogous to the ordinary law of exponents. The factorial power $x^{\overline{n}}$ is defined by
$x^{\overline{n}} = x(x+1)\cdots(x+n-1), \qquad n \ge 1, \quad x^{\overline{0}} = 1,$
and similarly $x^{\overline{m}} = x(x+1)\cdots(x+m-1)$. For ordinary exponents, the identity $x^{m+n} = x^m x^n$ relies on the commutativity of multiplication and the identical base $x$ across the factors. In contrast,
$x^{\overline{m+n}} = x(x+1)\cdots(x+m+n-1)$
contains $m+n$ consecutive integers starting from $x$, whereas
$x^{\overline{m}} , x^{\overline{n}} = \bigl(x(x+1)\cdots(x+m-1)\bigr) \bigl(x(x+1)\cdots(x+n-1)\bigr)$
does not produce the same sequence of factors unless $m=0$ or $n=0$. The two products differ because the second product repeats the initial $m$ factors of $x^{\overline{m}}$ and then continues with $x, x+1, \dots, x+n-1$, rather than continuing from $x+m$ as in $x^{\overline{m+n}}$. Therefore the factorial power law $x^{\overline{m+n}} = x^{\overline{m}} , x^{\overline{n}}$ fails in general.
This completes the proof.
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