Using Mathematical induction, prove that, for each $n \in \aleph$, $n < 3^n$.
I prove that using following steps. Please tell me my answer is correct or not?
$$n < 3^n$$ for $n = 0$, $n < 3^n \iff 0 < 3^0$
for $n = k$, $k < 3^k \iff 0 < 3^0$
for $n = k + 1$, $k < 3^k \iff k + 1 < 3^{k + 1}$
therefore $n < 3^n$
Clearly true for $k=0$, $k=1$.
Assume it true for $k$, i.e. $k < 3^k$.
Induction step for $k+1$ $$ \begin{align} k+1 &\le k + 2 k \\ &= 3 k \\ &< 3 ~3^k &&\hbox{Induction hypothesis}\\ &= 3^{k+1} \end{align} $$
Yes, it's clearly true. I mean, it's true for $k = 0$, $k = 1$ and from that point on $a = 3^k$ is going to result into $a$ getting multiplied by 3 for each increase of $k$. It also easily is to be seen in the formulas $f(x) = x$ and $g(x) = 3^x$. You're trying to prove that the domain $D_{g(x) > f(x)} = [0, \infty>$ and that is true (you can see it in a graph, for example). They also never intersect: $$f(x) = g(x)$$ $$x = 3^x$$ $$\log_3(x) = x = 3^x$$ Which never is true (except for $\lim_\limits{x \to \infty}$). That having said, it's impossible to get $k > 3^k$, so $k \leq 3^k$ with $$D_{g(x) > f(x)} = <-\infty, \infty>$$ (They do intersect at $\lim_\limits{x\to\infty}$ (not the negative one though), because $\lim_\limits{x\to-\infty}f(x) = -\infty$ and $\lim_\limits{x\to-\infty}g(x) = 0$, but $\lim_\limits{x\to\infty}f(x)= \infty$ and $\lim_\limits{x\to-\infty}g(x) = \infty$, however, I doubt highly if it's a natural number, as stated in your proof :p And if you still want to make it "for any $n$", just change "$<$" to "$\leq$").
So the results of your proof are correct. Congratz! The only one thing I have to say about it is that the notations aren't entirely correct, and may cause some confusion. However, your thinking steps were correct, and, to me at least, that is what matters most.