Metamath Proof Explorer


Theorem facgam

Description: The Gamma function generalizes the factorial. (Contributed by Mario Carneiro, 9-Jul-2017)

Ref Expression
Assertion facgam ( 𝑁 ∈ ℕ0 → ( ! ‘ 𝑁 ) = ( Γ ‘ ( 𝑁 + 1 ) ) )

Proof

Step Hyp Ref Expression
1 fveq2 ( 𝑥 = 0 → ( ! ‘ 𝑥 ) = ( ! ‘ 0 ) )
2 fv0p1e1 ( 𝑥 = 0 → ( Γ ‘ ( 𝑥 + 1 ) ) = ( Γ ‘ 1 ) )
3 gam1 ( Γ ‘ 1 ) = 1
4 2 3 syl6eq ( 𝑥 = 0 → ( Γ ‘ ( 𝑥 + 1 ) ) = 1 )
5 1 4 eqeq12d ( 𝑥 = 0 → ( ( ! ‘ 𝑥 ) = ( Γ ‘ ( 𝑥 + 1 ) ) ↔ ( ! ‘ 0 ) = 1 ) )
6 fveq2 ( 𝑥 = 𝑛 → ( ! ‘ 𝑥 ) = ( ! ‘ 𝑛 ) )
7 fvoveq1 ( 𝑥 = 𝑛 → ( Γ ‘ ( 𝑥 + 1 ) ) = ( Γ ‘ ( 𝑛 + 1 ) ) )
8 6 7 eqeq12d ( 𝑥 = 𝑛 → ( ( ! ‘ 𝑥 ) = ( Γ ‘ ( 𝑥 + 1 ) ) ↔ ( ! ‘ 𝑛 ) = ( Γ ‘ ( 𝑛 + 1 ) ) ) )
9 fveq2 ( 𝑥 = ( 𝑛 + 1 ) → ( ! ‘ 𝑥 ) = ( ! ‘ ( 𝑛 + 1 ) ) )
10 fvoveq1 ( 𝑥 = ( 𝑛 + 1 ) → ( Γ ‘ ( 𝑥 + 1 ) ) = ( Γ ‘ ( ( 𝑛 + 1 ) + 1 ) ) )
11 9 10 eqeq12d ( 𝑥 = ( 𝑛 + 1 ) → ( ( ! ‘ 𝑥 ) = ( Γ ‘ ( 𝑥 + 1 ) ) ↔ ( ! ‘ ( 𝑛 + 1 ) ) = ( Γ ‘ ( ( 𝑛 + 1 ) + 1 ) ) ) )
12 fveq2 ( 𝑥 = 𝑁 → ( ! ‘ 𝑥 ) = ( ! ‘ 𝑁 ) )
13 fvoveq1 ( 𝑥 = 𝑁 → ( Γ ‘ ( 𝑥 + 1 ) ) = ( Γ ‘ ( 𝑁 + 1 ) ) )
14 12 13 eqeq12d ( 𝑥 = 𝑁 → ( ( ! ‘ 𝑥 ) = ( Γ ‘ ( 𝑥 + 1 ) ) ↔ ( ! ‘ 𝑁 ) = ( Γ ‘ ( 𝑁 + 1 ) ) ) )
15 fac0 ( ! ‘ 0 ) = 1
16 oveq1 ( ( ! ‘ 𝑛 ) = ( Γ ‘ ( 𝑛 + 1 ) ) → ( ( ! ‘ 𝑛 ) · ( 𝑛 + 1 ) ) = ( ( Γ ‘ ( 𝑛 + 1 ) ) · ( 𝑛 + 1 ) ) )
17 facp1 ( 𝑛 ∈ ℕ0 → ( ! ‘ ( 𝑛 + 1 ) ) = ( ( ! ‘ 𝑛 ) · ( 𝑛 + 1 ) ) )
18 nn0p1nn ( 𝑛 ∈ ℕ0 → ( 𝑛 + 1 ) ∈ ℕ )
19 18 nncnd ( 𝑛 ∈ ℕ0 → ( 𝑛 + 1 ) ∈ ℂ )
20 eldifn ( ( 𝑛 + 1 ) ∈ ( ℤ ∖ ℕ ) → ¬ ( 𝑛 + 1 ) ∈ ℕ )
21 20 18 nsyl3 ( 𝑛 ∈ ℕ0 → ¬ ( 𝑛 + 1 ) ∈ ( ℤ ∖ ℕ ) )
22 19 21 eldifd ( 𝑛 ∈ ℕ0 → ( 𝑛 + 1 ) ∈ ( ℂ ∖ ( ℤ ∖ ℕ ) ) )
23 gamp1 ( ( 𝑛 + 1 ) ∈ ( ℂ ∖ ( ℤ ∖ ℕ ) ) → ( Γ ‘ ( ( 𝑛 + 1 ) + 1 ) ) = ( ( Γ ‘ ( 𝑛 + 1 ) ) · ( 𝑛 + 1 ) ) )
24 22 23 syl ( 𝑛 ∈ ℕ0 → ( Γ ‘ ( ( 𝑛 + 1 ) + 1 ) ) = ( ( Γ ‘ ( 𝑛 + 1 ) ) · ( 𝑛 + 1 ) ) )
25 17 24 eqeq12d ( 𝑛 ∈ ℕ0 → ( ( ! ‘ ( 𝑛 + 1 ) ) = ( Γ ‘ ( ( 𝑛 + 1 ) + 1 ) ) ↔ ( ( ! ‘ 𝑛 ) · ( 𝑛 + 1 ) ) = ( ( Γ ‘ ( 𝑛 + 1 ) ) · ( 𝑛 + 1 ) ) ) )
26 16 25 syl5ibr ( 𝑛 ∈ ℕ0 → ( ( ! ‘ 𝑛 ) = ( Γ ‘ ( 𝑛 + 1 ) ) → ( ! ‘ ( 𝑛 + 1 ) ) = ( Γ ‘ ( ( 𝑛 + 1 ) + 1 ) ) ) )
27 5 8 11 14 15 26 nn0ind ( 𝑁 ∈ ℕ0 → ( ! ‘ 𝑁 ) = ( Γ ‘ ( 𝑁 + 1 ) ) )