The result you're looking for is more general. In fact, it holds for any finite separable extension of fields. In particular, it holds for any finite extension of a perfect field (e.g. of $\mathbf{Q}$). I will sketch a proof below.
Claim: If $L/K$ is a finite separable extension of fields, then there is an element $\gamma\in L$ such that $L=K(\gamma)$.
Notation: If $x\in L$ and if $F$ is a subfield of $L$, then I will let $m_{x,F}$ denote the minimal polynomial of $x$ over $F$. Note that the minimal polynomial satisfies the following property: if $b\in L$ and if $P\in F[X]$ with $P(b)=0$, then $m_{x,F}|P$ in $F[X]$.
Proof: Suppose that $L/K$ is a finite separable extension of an infinite field $K$ (if $K$ is finite, the claim is easy to show) and let $\overline{L}=\overline{K}$ be an algebraic closure (one can quickly show that every finite extension of $\mathbf{Q}$ is separable; see Keith Conrad's notes on perfect fields). Since $L/K$ is finite, there are elements $a_1,\ldots, a_n\in L$ such that $L=K(a_1,\ldots, a_n)$. We can prove that $L=K(\alpha)$ for some $\alpha\in L$ by induction on the number of generators.
Let $F=K(a_1,\ldots, a_{n-2})$, so that $L=F(a_{n-1},a_n)$. Put $\alpha=a_{n-1}$ and $\beta=a_n$. Consider the finite set $S=\{\lambda\in K | \lambda=\frac{\alpha-\alpha'}{\beta'-\beta}\text{ with } \alpha', \beta'\in\overline{K} \text{ roots of }m_{\alpha,K}\text{ and }m_{\beta,K}\text{ resp., and }\beta'\neq\beta\}$ Let $\lambda\in K\backslash S$ (which exists since $K$ is infinite). Consider the sub-extension $L/F(\gamma)/K$ where $\gamma=\alpha+\lambda\beta$. Suppose that $L\neq F(\gamma)$; we will search for a contradiction. Note that if $\beta\in F(\gamma)$, then so is $\alpha$ (since $S$ contains $0$, $\lambda\neq 0$). We therefore must have $\deg m_{\beta,F(\gamma)}(X)\geq 2$. Consider the polynomial $m_{\alpha,K}(\gamma - \lambda X)\in F(\gamma)[X]$, which has $\beta$ as a root. By the general fact on minimal polynomials that I stated at the top, we have that $m_{\beta,F(\gamma)}|m_{\alpha,K}(\gamma - \lambda X)$ in $F(\gamma)[X]$. In particular, if $\beta'\neq \beta$ is a root of $m_{\beta,F(\gamma)}$ (which exists since $m_{\beta,F(\gamma)}$ is of degree $>1$), then $\beta'$ is also a root of $m_{\alpha,K}(\gamma - \lambda X)$; that is to say, the element $\alpha':=\gamma - \lambda\beta'$ is a root of $m_{\alpha,K}(X)$, and therefore $\lambda = \frac{\alpha - \alpha'}{\beta'-\beta}$ and $\lambda\in S$ (a contradiction).
Therefore, $L=F(\gamma)=K(a_1,\ldots,a_{n-2},\gamma)$ and we've reduced the number of generators of $L/K$ by $1$, go by induction now.