[proofplan] We verify the three independence axioms for the set system defining $M/e$. The nonloop hypothesis says precisely that $\{e\}$ is independent in $M$, which gives the empty independent set in $M/e$. Heredity and exchange are then inherited directly from the corresponding axioms in $M$ after adjoining the contracted element $e$. [/proofplan] [step:Verify that the empty set is independent after contraction] Since $e$ is a nonloop of $M$, the singleton set $\{e\}$ is independent in $M$, so $\{e\} \in \mathcal I$. Also $\varnothing \subset E \setminus \{e\}$ and $\varnothing \cup \{e\} = \{e\}$. By the definition of $\mathcal I(M/e)$, this implies $\varnothing \in \mathcal I(M/e)$. [/step] [step:Pass heredity from $M$ to $M/e$] Let $I \in \mathcal I(M/e)$, and let $J \subset I$. Since $I \subset E \setminus \{e\}$, also $J \subset E \setminus \{e\}$. By the definition of contraction, $I \cup \{e\} \in \mathcal I$. Since $J \subset I$, we have $J \cup \{e\} \subset I \cup \{e\}$. The hereditary axiom for the matroid $M$ therefore gives $J \cup \{e\} \in \mathcal I$. Applying the definition of $\mathcal I(M/e)$ again, we obtain $J \in \mathcal I(M/e)$. [guided] We need to prove the hereditary axiom for the contracted independence system. This means: whenever $I$ is independent in $M/e$ and $J$ is a subset of $I$, the set $J$ must also be independent in $M/e$. Start with $I \in \mathcal I(M/e)$ and $J \subset I$. The definition of contraction translates the statement $I \in \mathcal I(M/e)$ into the statement $I \cup \{e\} \in \mathcal I$. Because $J \subset I$, adjoining the same element $e$ preserves inclusion, so $J \cup \{e\} \subset I \cup \{e\}$. Now we use the hereditary axiom in the original matroid $M$: every subset of an independent set is independent. Since $I \cup \{e\}$ is independent in $M$, the subset $J \cup \{e\}$ is also independent in $M$. Finally, we translate back through the definition of contraction. Since $J \subset E \setminus \{e\}$ and $J \cup \{e\} \in \mathcal I$, the set $J$ belongs to $\mathcal I(M/e)$. Thus heredity holds for $M/e$. [/guided] [/step] [step:Pass exchange from $M$ to $M/e$] Let $I,J \in \mathcal I(M/e)$ satisfy $|I| < |J|$. By the definition of contraction, $I \cup \{e\} \in \mathcal I$ and $J \cup \{e\} \in \mathcal I$. Since $I,J \subset E \setminus \{e\}$, the element $e$ belongs to neither $I$ nor $J$, and hence $|I \cup \{e\}| = |I| + 1$ and $|J \cup \{e\}| = |J| + 1$. Thus $|I \cup \{e\}| < |J \cup \{e\}|$. Applying the exchange axiom in the matroid $M$ to the independent sets $I \cup \{e\}$ and $J \cup \{e\}$, there exists an element $x \in (J \cup \{e\}) \setminus (I \cup \{e\})$ such that $(I \cup \{e\}) \cup \{x\} \in \mathcal I$. Because $e \in I \cup \{e\}$, the element $x$ is not equal to $e$. Therefore $x \in J \setminus I$. Also $(I \cup \{x\}) \cup \{e\} = (I \cup \{e\}) \cup \{x\}$. Since the right-hand side is independent in $M$, the definition of contraction gives $I \cup \{x\} \in \mathcal I(M/e)$. This is exactly the exchange axiom for $M/e$. [/step] [step:Conclude that the contracted independence system is a matroid] The set system $\mathcal I(M/e)$ on the ground set $E \setminus \{e\}$ contains the empty set, is hereditary, and satisfies the exchange axiom. Therefore $M/e = (E \setminus \{e\}, \mathcal I(M/e))$ is a matroid. [/step]