If $k$ is the smallest integer such that $[a^k]>[a]^k$, which of the following is true? This is a question from the KVPY(SX)-2014 (an examination to get into various research institutes in India) paper.

Q. For a real number $r$ let $[r]$ denote the largest integer less than or equal to $r$. Let $a > 1$ be a real number which is not an integer and let $k$ be the smallest positive integer such that $[a^k] > [a]^k$. Then which of the following statements is always true?
$(A)\,\, k\le 2([a]+1)^2\quad (B) \,\, k\le ([a]+1)^4\quad (C)\,\, k\le 2^{[a]+1}\quad (D)\,\, k\le\frac{1}{a-[a]}+1$

The solutions which I’ve looked through on the Internet and on the booklet I’ve been given mostly state that by taking different values of a and k, option B is possible ( e.g.  here). But have not provided a proof on why. Or have put forward a proof that doesn't click my head (e.g. this).
I'm looking for a proof that suits the needs for a 12th grader in India.

You can argue for (D) by ruling out (A), (B), and (C), on the basis that they depend only on $[a]$, and not $[a]-a$. Any bound on $k$ depending only on $[a]$ can be defeated by choosing $a-[a]$ small enough. This kind of reasoning is quick and may be valuable on a timed test.
To see why (D) works (this part is unnecessary to answer the multiple choice question, but is a good sanity check if there is time), write $a = n+r$, where $n\in \mathbb{Z}$ and $0 < r < 1$. Put $k = \lceil 1/r \rceil$. We must show $(n+r)^k - n^k \ge 1$. By the binomial theorem,
$$
(n+r)^k - n^k = \sum_{i=1}^k \binom{k}{i} r^i n^{k-i} \ge k r n^{k-1} \ge n^{k-1} \ge 1
$$
