The proof that m is in fact the least fixed point can be done by showing that any element in \mathbb{M} is smaller than any fixed-point of f (because by property of supremum, if all elements of a set D \subseteq L are smaller than an element of L then also \sup(D) is smaller than that same element of L). This is done by induction: Assume k is some fixed-point of f. We now prove by induction over i that \forall i \in \mathbb{N}: f^i(\bot) \sqsubseteq k. The base of the induction (i = 0) obviously holds: f^0(\bot) = \bot \sqsubseteq k, since \bot is the least element of L. As the induction hypothesis, we may assume that f^i(\bot) \sqsubseteq k. We now do the induction step: From the induction hypothesis and the monotonicity of f (again, implied by the Scott-continuity of f), we may conclude the following: f^i(\bot) \sqsubseteq k ~\implies~ f^{i+1}(\bot) \sqsubseteq f(k). Now, by the assumption that k is a fixed-point of f, we know that f(k) = k, and from that we get f^{i+1}(\bot) \sqsubseteq k.
