(This is probably obvious but I will still make this obligatory remark since it wasn't mentioned before.)
Let $K=\Bbb Q(a_1,\dots,a_n)$. Since $\mathcal O_K=\bigcap_{\mathfrak p} \mathcal O_{K,\mathfrak p}$ it suffices to prove that $a_1,\dots,a_n$ have non-negative valuation at all primes $\mathfrak p$. Thus we look at all possible embeddings $K\to\Bbb C_p$ for all primes $p$ and look at $a_1,\dots,a_n$ in $\Bbb C_p$ (or $K_\mathfrak p$ instead of $\Bbb C_p$).
Lemma: Let $f(T)=\sum_{i=0}^\infty a_iT^i$ be a power series in $\Bbb C_p[[T]]$ (or over any valued field). Suppose that formally $f(T)=g(T)/h(T)$ where $g(T),h(T)\in\Bbb C_p[T]$ are coprime polynomials. If $f(T)$ is convergent with radius of convergence $R$, then $g(T)/h(T)$ is regular at $x$ and $f(x)=g(x)/h(x)$ for all $x$ with $|x|<R$.
Proof: We have $h(T)f(T)=g(T)$. It is a general result that evaluation and formal product of convergent power series commute (see e.g. Robert 'A Course in $p$-adic Analysis' Chapter 6, Section 1.2, Proposition 2), hence $h(x)f(x)=g(x)$ and it is not possible that $h(x)=0$ since $h$ and $g$ are coprime.
Now let $f(T)=n+(a_1+\dots+a_n)T+(a_1^2+\dots+a_n^2)T^2+\dots$. As formal series we have $f(T)=\sum_{i=1}^n \frac{1}{1-a_iT}=\frac{g(T)}{h(T)}$ where $g(T)=\sum_{i=1}^n\prod_{j\ne i}(1-a_jT)$, $h(T)=(1-a_1T)\dots(1-a_nT)$. Now, by assumption $f(T)\in\Bbb Z_p[[T]]$, thus $f$ has radius of convergence at least $1$, i.e. converges in $U=\{x\in\Bbb C_p\mid |x|<1\}$. By the lemma the rational function $g(T)/h(T)$ is regular in $U$. But notice that $a_i^{-1}$ will always be a pole of $g(T)/h(T)$ (if the $a_i$ are not distinct, $g(T)$ and $h(T)$ are not coprime but $(1-a_iT)$ always appears one more often in $h(T)$ than in $g(T)$), hence $a_i^{-1}\notin U$, i.e. $|a_i|\leq1$.
