Prove using induction : $n < 3^n$ $ p(n) = n < 3^n = q(n) $
when $n=1$, $p(n)=1< 3=q(n)$
Assume the result is true for $n=m$
$p(m)=m < 3^m$
when $n = m+1$
$p(m+1) = m+1 < 3^m +1<3*3^m = 3^{m+1}=q(m+1)$
is this correct?

I tried up to this extent to solve this. And also tried other methods. But no luck. Can some one show me the way to get this done.
Thank you.
 A: Yes, your proof correct. There is a more general way to proceed that lends further insight.
Hint $\  $ We seek an inductive proof of $\:\dfrac{3^n}n > 1.\,$ True for $\, n=1.\,$ For the inductive step, note
$$\quad \dfrac{3^{n+1}}{n+1}\, =\, \color{#0a0}{\dfrac{3\cdot n}{n+1}} \color{#c00}{\dfrac{3^n}n} > 1$$
since red $\rm\color{#c00}{term} > 1$ by induction, $ $ and the  green $\,\rm\color{#0a0}{term} > 1\,$ by $\, n>1 \Rightarrow 3n > n\! +\! 1.\ \ $ QED
Remark $\, $ Unwinding the induction yields a proof by multiplicative telescopy, e.g.
$$ \dfrac{3^5}5 =\, \dfrac{3\cdot\color{#c00}4}{5\quad\ } \dfrac{3\cdot\color{#0a0}3}{\color{#c00}4\quad\ } \dfrac{3\cdot\color{orange}2}{\color{#0a0}3\quad\ }\dfrac{3\cdot\color{brown}1}{\color{orange}2\quad\ }\dfrac{3}{\color{brown}1} $$
Note how same-color terms "telescopically" cancel, leaving five $3$'s on top, and $5$ on the bottom. Since each fraction is $> 1$ so  too is their product (a trivial induction). In essence, telescopy simplifies the induction to a trivial induction, that a product of terms $> 1$ is $> 1$. 
Similarly, telescopic methods often serve to greatly simply inductive proofs involving not only powers but also products, e.g. factorials. Similarly, there is also additive telescopy for sums.
A: For $n=1$ the statement is clear. Hence we assume that $n\leq 3^n$ holds for $n$ and we want to show the formula for $n+1$. 
$$n+1 \leq 3^n +1 \leq 3^n +3 \leq 3^{n+1}$$ Do you see why each estimate is true?
A: The correct answer is YES.  Your answer was correct and it seems people didn't actually read your post because it was messy. I've cleaned it up and it was clearly correct
A: For $n=1$ it is true and $1<3^1=3$.
Assume now that for a $k\ge 1$, we have that $k<3^k$. Then $3k<3\cdot 3^k=3^{k+1}$. But $3k>2k+1>k+1$, and altogether 
$$
k+1<3k<3^{k+1}.
$$
Thus the inductive hypothesis $\forall k\big(p(k)\Rightarrow P(k+1)\big)$ is valid, and so is the inequality.
