[proofplan] Write $M=(E,\mathcal I)$, where $E$ is the ground set and $\mathcal I$ is the independent-set family of $M$. By the definition of deletion, let $E' := E\setminus\{e\}$ denote the deleted ground set and let $\mathcal I' := \{I\in\mathcal I : I\subset E'\}$ denote the independent-set family of $M\setminus e$. We verify the three independence axioms for the set system $\mathcal I'$ on $E'$. Since $\mathcal I'$ consists exactly of the independent sets of $M$ that avoid $e$, the empty set and hereditary axioms follow from the corresponding axioms in $M$. For exchange, two independent sets in $\mathcal I'$ are independent in $M$, so the exchange axiom in $M$ supplies an augmenting element; because that element lies in the larger set, it cannot be $e$. [/proofplan]

[step:Verify that the empty set is independent after deletion] Retain the notation $M=(E,\mathcal I)$, $E' := E\setminus\{e\}$, and $\mathcal I' := \{I\in\mathcal I : I\subset E'\}$ from the proof plan. Since $M$ is a matroid, $\varnothing\in\mathcal I$. Also $\varnothing\subset E'$. Therefore $\varnothing\in\mathcal I'$ by the definition of $\mathcal I'$. [/step]

[step:Show that independence is inherited by subsets avoiding $e$] Let $I\in\mathcal I'$ and let $K\subset I$. Since $I\in\mathcal I'$, we have $I\subset E'$ and $I\in\mathcal I$. Hence $K\subset E'$. Since $\mathcal I$ is hereditary in $M$ and $K\subset I\in\mathcal I$, we have $K\in\mathcal I$. Therefore $K\in\mathcal I'$. [/step]

[step:Use exchange in $M$ and observe that the exchanged element is not $e$]Let $I,J\in\mathcal I'$ satisfy $|I|<|J|$. By the definition of $\mathcal I'$, both $I$ and $J$ belong to $\mathcal I$. Since $M$ is a matroid, the exchange axiom in $M$ gives an element $x\in J\setminus I$ such that $I\cup\{x\}\in\mathcal I$. Because $J\subset E'=E\setminus\{e\}$ and $x\in J$, we have $x\neq e$. Hence $I\cup\{x\}\subset E'$. Combining this with $I\cup\{x\}\in\mathcal I$, the definition of $\mathcal I'$ gives $I\cup\{x\}\in\mathcal I'$. Thus $\mathcal I'$ satisfies the exchange axiom.[/step]

[guided]We must prove the exchange axiom for the deleted set system $\mathcal I'$. So take two sets $I,J\in\mathcal I'$ with $|I|<|J|$. The definition of deletion says exactly that sets in $\mathcal I'$ are independent sets of $M$ that avoid the deleted element $e$. Therefore $I\in\mathcal I$, $J\in\mathcal I$, and both $I$ and $J$ are subsets of $E'=E\setminus\{e\}$. Now we may use the exchange axiom in the original matroid $M$. Since $I,J\in\mathcal I$ and $|I|<|J|$, there exists an element $x\in J\setminus I$ such that $I\cup\{x\}\in\mathcal I$. The only remaining point is to check that this new independent set still avoids the deleted element. Because $x\in J$ and $J\subset E\setminus\{e\}$, we have $x\neq e$. Also $I\subset E\setminus\{e\}$ because $I\in\mathcal I'$. Hence $I\cup\{x\}\subset E\setminus\{e\}$. Thus $I\cup\{x\}$ is independent in $M$ and is contained in the deleted ground set. By the definition of $\mathcal I'$, this means $I\cup\{x\}\in\mathcal I'$. This is precisely the exchange axiom for $M\setminus e$.[/guided]

[step:Conclude that the deleted set system is a matroid] The set system $\mathcal I'$ on $E'=E\setminus\{e\}$ contains $\varnothing$, is hereditary, and satisfies the exchange axiom. Therefore $M\setminus e=(E',\mathcal I')$ is a matroid. [/step]