Theorem expnnval

Description: Value of exponentiation to positive integer powers. (Contributed by Mario Carneiro, 4-Jun-2014)

	Ref	Expression
Assertion	expnnval	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ ) → ( 𝐴 ↑ 𝑁 ) = ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ 𝑁 ) )

Proof

Step	Hyp	Ref	Expression
1		nnz	⊢ ( 𝑁 ∈ ℕ → 𝑁 ∈ ℤ )
2		expval	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℤ ) → ( 𝐴 ↑ 𝑁 ) = if ( 𝑁 = 0 , 1 , if ( 0 < 𝑁 , ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ 𝑁 ) , ( 1 / ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ - 𝑁 ) ) ) ) )
3	1 2	sylan2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ ) → ( 𝐴 ↑ 𝑁 ) = if ( 𝑁 = 0 , 1 , if ( 0 < 𝑁 , ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ 𝑁 ) , ( 1 / ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ - 𝑁 ) ) ) ) )
4		nnne0	⊢ ( 𝑁 ∈ ℕ → 𝑁 ≠ 0 )
5	4	neneqd	⊢ ( 𝑁 ∈ ℕ → ¬ 𝑁 = 0 )
6	5	iffalsed	⊢ ( 𝑁 ∈ ℕ → if ( 𝑁 = 0 , 1 , if ( 0 < 𝑁 , ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ 𝑁 ) , ( 1 / ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ - 𝑁 ) ) ) ) = if ( 0 < 𝑁 , ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ 𝑁 ) , ( 1 / ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ - 𝑁 ) ) ) )
7		nngt0	⊢ ( 𝑁 ∈ ℕ → 0 < 𝑁 )
8	7	iftrued	⊢ ( 𝑁 ∈ ℕ → if ( 0 < 𝑁 , ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ 𝑁 ) , ( 1 / ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ - 𝑁 ) ) ) = ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ 𝑁 ) )
9	6 8	eqtrd	⊢ ( 𝑁 ∈ ℕ → if ( 𝑁 = 0 , 1 , if ( 0 < 𝑁 , ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ 𝑁 ) , ( 1 / ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ - 𝑁 ) ) ) ) = ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ 𝑁 ) )
10	9	adantl	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ ) → if ( 𝑁 = 0 , 1 , if ( 0 < 𝑁 , ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ 𝑁 ) , ( 1 / ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ - 𝑁 ) ) ) ) = ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ 𝑁 ) )
11	3 10	eqtrd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ ) → ( 𝐴 ↑ 𝑁 ) = ( seq 1 ( · , ( ℕ × { 𝐴 } ) ) ‘ 𝑁 ) )