Riddle : Given an integer power, does any integer to this power start with the power? I initially found the following riddle somewhere : Is there an integer $n$ such that $n^{2004}$ starts (from the left) by $2004$ ?
I was unable to find an answer, but I found the question rather interesting and couldn't find any such question on MSE, hence I'm wondering :
Given a non trivial positive integer $a$, when is there a positive integer $n$ such that $n^a$ starts by $a$ ?
(By trivial integers I mean cases like $a=1,a=2,\dots$ etc for which $n$ can very easily be found)

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 $k$.
