It is enough to show that \(\mu (\mathfrak {m}) = \dim \mathfrak {m} / \mathfrak {m}^{2}\), where \(\mu (\mathfrak {m})\) is the minimal number of generators of \(\mathfrak {m}\).

First, we show \(\mu (\mathfrak {m}) \leq \dim \mathfrak {m} / \mathfrak {m}^{2}\). Let \(\overline{x}_{1}, \ldots , \overline{x}_{n}\) be a basis of \(\mathfrak {m} / \mathfrak {m}^{2}\) over \(k\). Then, by a corollary of Nakayama’s lemma (Atiyah-MacDonald 2.8), the lifts of the \(\overline{x}_{i}\) generate \(\mathfrak {m}\), so \(\mu (\mathfrak {m}) \leq n\) as desired.

Second, we show that \(\dim \mathfrak {m} / \mathfrak {m}^{2} \leq \mu (\mathfrak {m})\). Let \(x_{1}, \ldots , x_{n}\) be a generating set of \(\mathfrak {m}\). Then, the residues of the \(x_{i}\) generate \(\mathfrak {m} / \mathfrak {m}^{2}\) (quotient map is a homomorphism or something). Since it is a generating set, it contains a basis by linear algebra, and \(\dim \mathfrak {m} / \mathfrak {m}^{2} \leq n\) as desired.