Prove that $A+I_n$ is invertible, where $\left\lVert A\right\rVert<1$ 
Let $\left\lVert\cdot\right\rVert:\mathbb R^{n\times n}\to\mathbb R$ be a submultiplicative matrix norm and $A\in\mathbb R^{n\times n}$ such that $\lVert A\rVert<1$. Prove that $A+I_n$ is invertible, where $I_n$ is the identity matrix in $\mathbb R^{n\times n}$.

I tried coming up with something like $$\lVert A+I_n\rVert=\lVert A(I_n+A^{-1})\rVert\leq\lVert A\rVert\cdot\lVert(I_n+A^{-1})\rVert<\lVert I_n+A^{-1}\rVert,$$
but that doesn't seem to get me anywhere. In the end, I think I should have some (in)equality with the determinant of $I_n+A^{-1}$ in it (and conclude that it is not $0$), but I don't know how to get there. How could I proceed?
 A: Note that for any matrix $M$: if $\lambda$ is an eigenvalue, then $|\lambda| < \|M\|$.  Thus, all eigenvalues of $A$ satisfy $|\lambda| < 1$.
Now, if $\mu$ is an eigenvalue of $A + I$, then $(\mu - 1)$ is an eigenvalue of $A + I$, which tells us that $|\mu - 1| < 1$.  We can conclude that $A + I$ does not have zero as an eigenvalue.  It follows that $A+I$ is invertible.
A: Given a matrix such that $\lVert \mathbf{A} \rVert < 1$, the matrix $\mathbf{I} \color{red}{-} \mathbf{A}$ is nonsingular with
$$
\left( \mathbf{I} - \mathbf{A} \right)^{-1} = \sum_{k=0}^{\infty}\mathbf{A}^{k},
$$
and
$$
 \lVert \left( \mathbf{I} \color{red}{-} \mathbf{A} \right)^{-1} \rVert
\le
\frac{1}{1-\lVert \mathbf{A}\rVert}.
$$
Proof by contradiction
Let $\mathbf{I} - \mathbf{A}$ be singular. $\exists$ a nonzero $x$ such that $\left( \mathbf{I} - \mathbf{A} \right)x = 0.$ Then we have
$$
 \lVert x \rVert = \lVert \mathbf{A} x \rVert
$$
which implies $\lVert \mathbf{A} \rVert \ge 1.$ $\color{red}{\Rightarrow \Leftarrow}$
Derivation
Start with the telescopic identity
$$
\left( 
\sum_{k=0}^{N}\mathbf{A}^{k}
\right)
%
\left( 
\mathbf{I} - \mathbf{A}
\right)
%
=
%
\mathbf{I} - \mathbf{A}^{N+1}
%
$$
Knowing the property of submultiplicative norms $\lVert \mathbf{A}^{k} \rVert \le \lVert \mathbf{A} \rVert^{k}$
and given $\lVert \mathbf{A} \rVert < 1$ we see $\lim_{k\to\infty}\mathbf{A}^{k} = 0$. This implies
$$
\left( 
\lim_{N\to \infty}
 \sum_{k=0}^{N}\mathbf{A}^{k}
\right)
%
\left( 
\mathbf{I} - \mathbf{A}
\right)
%
=
%
\mathbf{I},
%
$$
and 
$$
\left( 
\mathbf{I} - \mathbf{A}
\right)^{-1}
  =
\left( 
\lim_{N\to \infty}
 \sum_{k=0}^{N}\mathbf{A}^{k}
\right).
$$
At last,
$$
\lVert \left( 
\mathbf{I} - \mathbf{A}
\right)^{-1}
\rVert
  \le
 \sum_{k=0}^{\infty}\lVert \mathbf{A} \rVert^{k}
  =
\frac{1}{1-\lVert \mathbf{A}\rVert}
$$
A: A bit of a more straightforward proof, similar to @dantopa. 
Let $K = -A$, then $I + A = I - K$ and $\lVert K \rVert = \lVert A \rVert < 1$. Suppose for the sake of contradiction that $(I-K)$ is not invertible, then there exists $x\in\mathbb{R}^n\setminus\{0\}$, such that
    $$
  \lVert(I-K)x\rVert = 0, \lVert x \rVert \neq 0.
 $$
    But
    $$
  \lVert(I-K)x\rVert\geq \lVert x \rVert - \lVert Kx\rVert \geq \lVert x\rVert(1 - \lVert K \rVert) > 0,
 $$
    a contradiction, so $(I-K)$ is invertible.
Now let $S = (I-K)^{-1}$ and
    $$
  S_n = I + K + \cdots + K^n,
 $$
    for some $n\in\mathbb{N}$,
    then
    \begin{align*}
  KS_n = K + K^2 + \cdots + K^{n+1} = S_n + K^{n+1} - I,\\
  S_n= (I - K)^{-1}(I - K^{n+1}) = S(I-K^{n+1}),
 \end{align*}
    and so
    $$
  \lVert S_n - S\rVert = \lVert S(I - K^{n+1})-S\rVert = \lVert-SK^{n+1}\rVert \leq\lVert S\rVert\lVert{K\rVert}^{n+1} \xrightarrow{n\to\infty} 0.
 $$
Q.E.D.
