Metamath Proof Explorer


Theorem bcn0

Description: N choose 0 is 1. Remark in Gleason p. 296. (Contributed by NM, 17-Jun-2005) (Revised by Mario Carneiro, 8-Nov-2013)

Ref Expression
Assertion bcn0 ⊢ N ∈ ℕ 0 → ( N 0 ) = 1

Proof

Step Hyp Ref Expression
1 0elfz ⊢ N ∈ ℕ 0 → 0 ∈ 0 … N
2 bcval2 ⊢ 0 ∈ 0 … N → ( N 0 ) = N ! N − 0 ! ⁢ 0 !
3 1 2 syl ⊢ N ∈ ℕ 0 → ( N 0 ) = N ! N − 0 ! ⁢ 0 !
4 nn0cn ⊢ N ∈ ℕ 0 → N ∈ ℂ
5 4 subid1d ⊢ N ∈ ℕ 0 → N − 0 = N
6 5 fveq2d ⊢ N ∈ ℕ 0 → N − 0 ! = N !
7 fac0 ⊢ 0 ! = 1
8 oveq12 ⊢ N − 0 ! = N ! ∧ 0 ! = 1 → N − 0 ! ⁢ 0 ! = N ! ⋅ 1
9 6 7 8 sylancl ⊢ N ∈ ℕ 0 → N − 0 ! ⁢ 0 ! = N ! ⋅ 1
10 faccl ⊢ N ∈ ℕ 0 → N ! ∈ ℕ
11 10 nncnd ⊢ N ∈ ℕ 0 → N ! ∈ ℂ
12 11 mulridd ⊢ N ∈ ℕ 0 → N ! ⋅ 1 = N !
13 9 12 eqtrd ⊢ N ∈ ℕ 0 → N − 0 ! ⁢ 0 ! = N !
14 13 oveq2d ⊢ N ∈ ℕ 0 → N ! N − 0 ! ⁢ 0 ! = N ! N !
15 facne0 ⊢ N ∈ ℕ 0 → N ! ≠ 0
16 11 15 dividd ⊢ N ∈ ℕ 0 → N ! N ! = 1
17 14 16 eqtrd ⊢ N ∈ ℕ 0 → N ! N − 0 ! ⁢ 0 ! = 1
18 3 17 eqtrd ⊢ N ∈ ℕ 0 → ( N 0 ) = 1