sequences and series - Show that $a_n^{1/n}$ converges as $ntoinfty$

Suppose that a sequence ${a_n : n = 1,2,\cdots}$ real numbers is such that
$a_n\geq 1$ for all $n ≥ 1$ and

$$a_{n+m}\leq a_na_m\quad \rm{for~ all}~n\geq1, m\geq1.$$

Show that $a_n^{1/n}$ converges as $n\to\infty$

My solution:
By taking log, we have

$$\log{a_{n+m}}\leq \log{a_na_m}=\log{a_n}+\log{a_m}\quad \rm{for~ all}~n\geq1, m\geq1.$$

So we have

$$\begin{align} \log{a_2}&\leq \log{a_1}+\log{a_1}=2\log{a_1} \\ \log{a_3}&\leq \log{a_1}+\log{a_2}\leq3\log{a_1} \\ \cdots \\ \log{a_n}&\leq n\log{a_1} \end{align}$$
So,
$$\log{a_n}^{1/n}=\frac{1}{n}\log{a_n}$$
So I can prove $\log a_n$ is bounded but cannot to prove it's monotonic which can sufficiently lead to $\log a_n$ converges. How to deal with it?

Or if we cannot prove monotonic, how to prove the limit exists?