Riddle : Given an integer power, does any integer to this power start with the power?

Yes, this is always possible. As stated in the comments, you are looking for an $n\in\mathbb{N}$ such that $a\cdot 10^k\leq n^a<(a+1)\cdot 10^k$ for some $k\in\mathbb{N}$. For any $k$, let $m_k$ be the largest integer such that $m_k^a< a\cdot 10^k$. Then $(m_k+1)^a\geq a\cdot 10^k$, and $(m_k+1)^a<(a+1)\cdot 10^k$ as long as $$\left(\frac{m_k+1}{m_k}\right)^a\leq\frac{a+1}{a}.$$ This last inequality is true as long as $m_k$ is sufficiently large, since the left-hand side converges to $1$ as $m_k\to \infty$. But $m_k\to\infty$ as $k\to\infty$, so this inequality must hold for all sufficiently large $k$. Thus $n=m_k+1$ is a number of the sort you seek for all sufficiently large $k$.

This is a special case of the following theorem (Lemma 2.1 in this paper on disjunctive sequences), whose proof is along the same lines as that posted in the answer by @EricWofsey:

If $a_1, a_2, a_3, \dots$ is a strictly increasing infinite sequence of positive integers such that $$\lim_{n\to \infty} \frac{a_{n+1}}{a_n} = 1$$ then for any positive integer $m$ and any integer base $b \ge 2$, there is an $a_n$ whose expression in base $b$ starts with the expression of $m$ in base $b$.

Your result is then the very special case of taking $a_n = n^k$ and $k=m$. For this and some other special cases, see these examples of disjunctive sequences. (E.g., for any desired positive integer, there are infinitely many prime numbers whose representation begins with the digits of that number.)

NB: For any positive integer exponent $k$ and any desired positive integer $m$, there are infinitely many positive integers $n$ such that the representation of $n^k$ starts with the representation of $m$; furthermore, this holds for digital representations in any integer base $b \ge 2$.