Prove that a power of odd number is always odd by induction. The problem has confused me for like half hour.
An integer is odd if it can be written as d = 2m+1. Use induction to prove that the ${d^n}$ = 1 (mod 2)
by induction, the basecase is pretty simple , let n = 0 then ${d^0}$=1 (mod 2) is correct. But I stucked in the I.H and inductive steps.
Any hints please? 
Thank you in advance!

It's easier to just show that the product of a finite number of odd integers is odd. This can be done inductively if you like. Then your problem is a special case of this.
Base case: a single odd number is odd.
Inductive step: Assume $n_1,n_2,\ldots,n_{k+1}$ are odd and that $p_k = (n_1)(n_2)\cdots(n_k)$ is odd. Then $n_{k+1}=2a+1$ and $p_k=2b+1$ for some integers $a,b$. Thus $p_{k+1}=p_kn_{k+1}=(2a+1)(2b+1)=4ab+2a+2b+1=2(2ab+a+b)+1$ is odd.
Hence the product of a finite number of odd integers is odd, and in particular $(2a+1)^k$ is odd for integral $a$ and positive integral $k$.
