We have
$$a + \frac{1}{a} = [a] + \left[ \frac{1}{a} \right] + \{ a \} + \left\{ \frac{1}{a} \right\} = [a] + \left[ \frac{1}{a} \right] + 1$$
is an integer and denote this integer by $k$. Let $S_{n} = a^{n} + \frac{1}{a^{n}}$, $n = 0, 1, 2, \ldots$ Since $a$ and $\frac{1}{a}$ are the roots of quadratic equation $x^{2} - kx + 1 = 0$, it follows that
$$S_{n+2} = k S_{n+1} - S_{n}, \quad n = 0, 1, 2, \ldots$$
We have $S_{0} = 2$, $S_{1} = k$ hence, by induction of step 2, we obtain that $S_{n} \in \mathbb{Z}$ for $n = 2, 3, \ldots$
It follows
$$\left\{ a^{n} \right\} + \left\{ \frac{1}{a^{n}} \right\} = a^{n} + \frac{1}{a^{n}} - \left[ a^{n} \right] - \left[ \frac{1}{a^{n}} \right] = S_{n} - \left[ a^{n} \right] - \left[ \frac{1}{a^{n}} \right] \in \mathbb{Z},$$
that is $\left\{ a^{n} \right\} + \left\{ \frac{1}{a^{n}} \right\} \in \mathbb{Z}$. We get $\left\{ a^{n} \right\} + \left\{ \frac{1}{a^{n}} \right\} \in \{ 0, 1 \}$. If $\left\{ a^{n} \right\} + \left\{ \frac{1}{a^{n}} \right\} = 0$, then $a^{n}$ and $\frac{1}{a^{n}}$ are both integers, not possible. Therefore, $\left\{ a^{n} \right\} + \left\{ \frac{1}{a^{n}} \right\} = 1$, and we are done.