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Take arbitrary `x`; we proceed by induction over `n`.

We have `x`^{0} = 1 = (**prod**_{1 <= i <= 0} `x`) and
thus the induction base holds.

We take arbitrary `n` in **N** and assume

(1)x^{n}= (prod_{1 <= i <= n}x).

We have to prove

(2)x^{n+1}= (prod_{1 <= i <= n+1}x).

We know

x^{n+1}= (definition exponentiation) x*x^{n}= (1) x* (prod_{1 <= i <= n}x)= (definition ( prod_{}))( prod_{1 <= i <= n+1}x)

which implies (2).

Author: Wolfgang Schreiner

Last Modification: November 24, 1999