Well, this should be obvious since an odd times an odd gives an odd, and all odd's are 1 (mod 2) (pretty much the definition of an odd number).
So you have the base case, now you just need to prove the inductive step:
Inductive hypothesis:
\begin{align*}
d^{k - 1} = 1 \text{ (mod 2)}
\end{align*}
Show that $d^k = d\cdot d^{k - 1} = 1 \text{ (mod }2\text{)}$. You already know that $d = 1$ (mod 2) (because $d$ is odd) and that $d^{k - 1} = 1$ (mod 2). Write out explicitly what this means:
\begin{align*}
d =& 2\lambda_1 + 1, d^{k - 1} = 2\lambda_2 + 1 \\
d\cdot d^{k - 1} =& (2\lambda_1 + 1)(2\lambda_2 + 1) = 4\lambda_1\lambda_2 + 2(\lambda_1 + \lambda_2) + 1 = 2\Lambda + 1 \\
d\cdot d^{k - 1} =& d^k = 1 \text{ (mod 2), q.e.d.}
\end{align*}
You already proved the base case and this proves the inductive step: if $d^{k - 1} = 1$ (mod 2) and $d = 1$ (mod 2) (i.e. $d$ is odd), then $d^k = 1$ (mod 2).