Metamath Proof Explorer


Theorem risefacval2

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

Ref Expression
Assertion risefacval2 ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → A N ‾ = ∏ k = 1 N A + k - 1

Proof

Step Hyp Ref Expression
1 risefacval ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → A N ‾ = ∏ n = 0 N − 1 A + n
2 1zzd ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → 1 ∈ ℤ
3 0zd ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → 0 ∈ ℤ
4 nn0z ⊢ N ∈ ℕ 0 → N ∈ ℤ
5 peano2zm ⊢ N ∈ ℤ → N − 1 ∈ ℤ
6 4 5 syl ⊢ N ∈ ℕ 0 → N − 1 ∈ ℤ
7 6 adantl ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → N − 1 ∈ ℤ
8 simpl ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → A ∈ ℂ
9 elfznn0 ⊢ n ∈ 0 … N − 1 → n ∈ ℕ 0
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 ∈ ℕ 0 ∧ 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 ∈ ℕ 0 → ∏ 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 ∈ ℕ 0 → 0 + 1 = 1
17 nn0cn ⊢ N ∈ ℕ 0 → N ∈ ℂ
18 1cnd ⊢ N ∈ ℕ 0 → 1 ∈ ℂ
19 17 18 npcand ⊢ N ∈ ℕ 0 → N - 1 + 1 = N
20 19 adantl ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → N - 1 + 1 = N
21 16 20 oveq12d ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → 0 + 1 … N - 1 + 1 = 1 … N
22 21 prodeq1d ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → ∏ k = 0 + 1 N - 1 + 1 A + k - 1 = ∏ k = 1 N A + k - 1
23 1 14 22 3eqtrd ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → A N ‾ = ∏ k = 1 N A + k - 1