And by substituting (n+1) for n in 1.03,
k^{(n+1)}=k(k-1)...(k-n+1)(k-n) [1.04]

Therefore:
k^{(n+1)}=k^{(n)}(k-n)

Values
for 0 and 1

Substituting 0 for n in 1.02 we find:
[1.05]
And substituring 1 for n:
[1.06]

Expression
for Negative Factorials

We have shown that k^{(0)} is 1, and we make the
following claim:
[1.07]
By induction, we claim that if 1.07 is true then so is (substituting
(-n+1 for -n)
[1.08]

We recall that
[1.09]
And
[1.10]

Hence, by substituting these in 1.07 and 1.08
[1.11]

[1.12]
From 1.11 and 1.12 we note:

Which is true by definition (1.02). Hence because 1.07 is true for n=0,
it is true for n=-1, and so true for all n.

Examples
for n=-1, -2, -3

If n=(-1), using
k^{(-1)}=1/(k+1)^{1}
k^{(-1)}=1/(k+1)

For n=(-2),
k^{(-2)}=1/(k+2)^{2}
k^{(-2)}=1/(k+2)(k+1)

For n=(-3),
k^{(-3)}=1/(k+3)^{3}
k^{(-3)}=1/(k+3)(k+2)(k+1)

Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle: