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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 ∈ 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 ) ) )