Proof by induction that $3^n - 1$ is an even number How to demonstrate that  $3^n - 1$ is an even number using the principle of induction? 
I tried taking that $3^k - 1$ is an even number and as a thesis I must demonstrate that $3^{k+1} - 1$ is an even number, but I can't make a logical argument. 
So I think it's wrong the assumption... 
Thanks

Hint $\ $ Let $\,a = 2\,$ below (= first term of binomial expansion).
$\ \ \ \begin{align} (1+ a)^n\, \ \  =&\,\ \ 1 + ak\qquad\qquad\quad {\rm i.e.}\ \ P(n)\\[1pt]
\Rightarrow\ (1+a)^{\color{#c00}{n+1}}\! =&\  (1+ak)(1 + a)\\[2pt]  
 =&\,\ \ 1+ a\,(\underbrace{k\!+\!1\!+\!ka\!}_{\large k'})\ \ \ {\rm i.e.}\ \ P(\color{#c00}{n\!+\!1})\\  
  \end{align}$
Remark $\ $ If you know modular arithmetic (congruences) then you can view it as a special case of the Congruence Power Rule, i.e. $\, x\equiv 1\,\Rightarrow\, x^n\equiv 1^n\equiv 1\pmod{\!a},\,$ where the induction is conceptually clearer: $ $ the powering of a congruence, and the trivial induction $\,1^n\equiv 1.$
