Prove that the inverse operator of (Id-M) is both bounded and has operator norm less than $\frac{1}{{1 - \gamma }}$

Suppose Id denotes the identity operator defined by $Id(v)=v$ for all $v \in V$ where V is the normed vector space $V$. Let also $M$ denote the operator $M:V \to V$ satisfying $\left\| M \right\| \le \gamma $ for some $\gamma$ less than 1 and greater than zero. I want to show that if $(Id-M):V\to V $ is invertible, then the inverse operator $(Id-M)^{-1}$ is bounded and $\left\| {\left( {Id - M} \right)} \right\| \le \frac{1}{{1 - \gamma }}$.

My answer: The first part follows from the open mapping theorem. Given a bijection, it maps open sets to open sets. Using this I could prove the inverse does same and hence bounded. However, I could not prove the second part (the bound on its norm).

Hint: $Id+M+M^{2}+\cdots$ converges in operator norm. If we denote this sum by $T$ then $T(Id-M)=(Id-M)T=I$. So $\|(Id-M)^{-1}|| =\|T\| \leq \sum_n \|M^{n}\| \leq \sum_n \gamma ^{n}=\frac 1 {1-\gamma}$.