let n be Element of NAT ; :: thesis: n !c = n !
defpred S₁[ Element of NAT ] means $1 !c = $1 ! ;
A1: S₁[ 0 ] by Th1, NEWTON:18;
A2: for k being Element of NAT st S₁[k] holds
S₁[k + 1]

proof

let k be Element of NAT ; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A3: k !c = k ! ; :: thesis: S₁[k + 1]
thus (k + 1) !c = (k ! ) * (k + 1) by A3, Th1
.= (k + 1) ! by NEWTON:21 ; :: thesis: verum

end;

A4: for k being Element of NAT holds S₁[k] from NAT_1:sch 1(A1, A2);
thus n !c = n ! by A4; :: thesis: verum