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Metamath Proof Explorer

Theorem expm1t

Description: Exponentiation in terms of predecessor exponent. (Contributed by NM, 19-Dec-2005)

Ref Expression
Assertion expm1t ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ ) → ( 𝐴 ↑ 𝑁 ) = ( ( 𝐴 ↑ ( 𝑁 − 1 ) ) · 𝐴 ) )

Proof

Step Hyp Ref Expression
1 nncn ⊢ ( 𝑁 ∈ ℕ → 𝑁 ∈ ℂ )
2 ax-1cn ⊢ 1 ∈ ℂ
3 npcan ⊢ ( ( 𝑁 ∈ ℂ ∧ 1 ∈ ℂ ) → ( ( 𝑁 − 1 ) + 1 ) = 𝑁 )
4 1 2 3 sylancl ⊢ ( 𝑁 ∈ ℕ → ( ( 𝑁 − 1 ) + 1 ) = 𝑁 )
5 4 oveq2d ⊢ ( 𝑁 ∈ ℕ → ( 𝐴 ↑ ( ( 𝑁 − 1 ) + 1 ) ) = ( 𝐴 ↑ 𝑁 ) )
6 5 adantl ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ ) → ( 𝐴 ↑ ( ( 𝑁 − 1 ) + 1 ) ) = ( 𝐴 ↑ 𝑁 ) )
7 nnm1nn0 ⊢ ( 𝑁 ∈ ℕ → ( 𝑁 − 1 ) ∈ ℕ0 )
8 expp1 ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝑁 − 1 ) ∈ ℕ0 ) → ( 𝐴 ↑ ( ( 𝑁 − 1 ) + 1 ) ) = ( ( 𝐴 ↑ ( 𝑁 − 1 ) ) · 𝐴 ) )
9 7 8 sylan2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ ) → ( 𝐴 ↑ ( ( 𝑁 − 1 ) + 1 ) ) = ( ( 𝐴 ↑ ( 𝑁 − 1 ) ) · 𝐴 ) )
10 6 9 eqtr3d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ ) → ( 𝐴 ↑ 𝑁 ) = ( ( 𝐴 ↑ ( 𝑁 − 1 ) ) · 𝐴 ) )