Theorem risefacval2

Description: One-based value of rising factorial. (Contributed by Scott Fenton, 15-Jan-2018)

	Ref	Expression
Assertion	risefacval2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) → ( 𝐴 RiseFac 𝑁 ) = ∏ 𝑘 ∈ ( 1 ... 𝑁 ) ( 𝐴 + ( 𝑘 − 1 ) ) )

Proof

Step	Hyp	Ref	Expression
1		risefacval	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) → ( 𝐴 RiseFac 𝑁 ) = ∏ 𝑛 ∈ ( 0 ... ( 𝑁 − 1 ) ) ( 𝐴 + 𝑛 ) )
2		1zzd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) → 1 ∈ ℤ )
3		0zd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) → 0 ∈ ℤ )
4		nn0z	⊢ ( 𝑁 ∈ ℕ0 → 𝑁 ∈ ℤ )
5		peano2zm	⊢ ( 𝑁 ∈ ℤ → ( 𝑁 − 1 ) ∈ ℤ )
6	4 5	syl	⊢ ( 𝑁 ∈ ℕ0 → ( 𝑁 − 1 ) ∈ ℤ )
7	6	adantl	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) → ( 𝑁 − 1 ) ∈ ℤ )
8		simpl	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) → 𝐴 ∈ ℂ )
9		elfznn0	⊢ ( 𝑛 ∈ ( 0 ... ( 𝑁 − 1 ) ) → 𝑛 ∈ ℕ0 )
10	9	nn0cnd	⊢ ( 𝑛 ∈ ( 0 ... ( 𝑁 − 1 ) ) → 𝑛 ∈ ℂ )
11		addcl	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑛 ∈ ℂ ) → ( 𝐴 + 𝑛 ) ∈ ℂ )
12	8 10 11	syl2an	⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) ∧ 𝑛 ∈ ( 0 ... ( 𝑁 − 1 ) ) ) → ( 𝐴 + 𝑛 ) ∈ ℂ )
13		oveq2	⊢ ( 𝑛 = ( 𝑘 − 1 ) → ( 𝐴 + 𝑛 ) = ( 𝐴 + ( 𝑘 − 1 ) ) )
14	2 3 7 12 13	fprodshft	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) → ∏ 𝑛 ∈ ( 0 ... ( 𝑁 − 1 ) ) ( 𝐴 + 𝑛 ) = ∏ 𝑘 ∈ ( ( 0 + 1 ) ... ( ( 𝑁 − 1 ) + 1 ) ) ( 𝐴 + ( 𝑘 − 1 ) ) )
15		0p1e1	⊢ ( 0 + 1 ) = 1
16	15	a1i	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) → ( 0 + 1 ) = 1 )
17		nn0cn	⊢ ( 𝑁 ∈ ℕ0 → 𝑁 ∈ ℂ )
18		1cnd	⊢ ( 𝑁 ∈ ℕ0 → 1 ∈ ℂ )
19	17 18	npcand	⊢ ( 𝑁 ∈ ℕ0 → ( ( 𝑁 − 1 ) + 1 ) = 𝑁 )
20	19	adantl	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) → ( ( 𝑁 − 1 ) + 1 ) = 𝑁 )
21	16 20	oveq12d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) → ( ( 0 + 1 ) ... ( ( 𝑁 − 1 ) + 1 ) ) = ( 1 ... 𝑁 ) )
22	21	prodeq1d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) → ∏ 𝑘 ∈ ( ( 0 + 1 ) ... ( ( 𝑁 − 1 ) + 1 ) ) ( 𝐴 + ( 𝑘 − 1 ) ) = ∏ 𝑘 ∈ ( 1 ... 𝑁 ) ( 𝐴 + ( 𝑘 − 1 ) ) )
23	1 14 22	3eqtrd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) → ( 𝐴 RiseFac 𝑁 ) = ∏ 𝑘 ∈ ( 1 ... 𝑁 ) ( 𝐴 + ( 𝑘 − 1 ) ) )