let n be even Nat; :: thesis: (- 1) |^ n = 1
defpred S1[ Nat] means ( $1 is even implies (- 1) |^ $1 = 1 );
A1: now :: thesis: for k being Nat st ( for n being Nat st n < k holds
S1[n] ) holds
S1[k]
let k be Nat; :: thesis: ( ( for n being Nat st n < k holds
S1[n] ) implies S1[b1] )

assume A2: for n being Nat st n < k holds
S1[n] ; :: thesis: S1[b1]
per cases ( k is odd or k is even ) ;
suppose A3: k is even ; :: thesis: S1[b1]
now :: thesis: ( ( k = 0 & S1[k] ) or ( k > 0 & S1[k] ) )
per cases ( k = 0 or k > 0 ) ;
case A4: k > 0 ; :: thesis: S1[k]
( 0 is even & 0 + 1 = 1 ) ;
then k - 2 in NAT by A3, A4, NAT_1:23, INT_1:5;
then reconsider k2 = k - 2 as Nat ;
A5: k2 + 2 = k ;
then A6: k2 is even by A3;
(- 1) |^ k = (- 1) |^ (k2 + 2)
.= ((- 1) |^ k2) * ((- 1) |^ 2) by NEWTON:8
.= 1 * ((- 1) |^ (1 + 1)) by A2, A6, A5, NAT_1:16
.= ((- 1) |^ 1) * (- 1) by NEWTON:6
.= (- 1) * (- 1) ;
hence S1[k] ; :: thesis: verum
end;
end;
end;
hence S1[k] ; :: thesis: verum
end;
end;
end;
for k being Nat holds S1[k] from NAT_1:sch 4(A1);
hence (- 1) |^ n = 1 ; :: thesis: verum