Question 1194395
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I think your base case and hypothesis are fine.  You could have made the step case (m+1) clearer, as follows:

base:  n=2, {{{ 3^2 = 9 }}} > {{{ 2^(2+1) = 8 }}}   base case n=2 holds
hypothesis:  Assume {{{3^n > 2^(n+1) }}} for some m, where m>2.

step case:  Let n=m+1:

      Need to show {{{ 3^(m+1) > 2^((m+1)+1) }}}

  LHS:  {{{ 3^(m+1) = 3*3^(m) }}}
  RHS:  {{{ 2^((m+1)+1) = 2*2^(m+1) }}}

We now apply the hypothesis {{{ 3^(m) > 2^(m+1) }}}

and can conclude LHS > RHS, i.e.  {{{ 3*(3^(m)) > 2*(2^(m+1)) }}}  
(for A > 0 and B > 0, if A > B than certainly 3A > 2B).

or, re-writing,  {{{ 3^(m+1) > 2^((m+1)+1) }}}

Done

Usually, in inductive proofs, you should be looking to reduce the 'n+1' case to something involving 'n'  so that you can directly apply the hypothesis, then include that new information into the 'n+1' scenario and draw the final conclusion.