Prove $(I-J_n)^{-1}=I-\frac{1}{n-1}J_n$ I have a question on matrices from a recent problem.

Let $J_n$ be the $n\times n$ matrix each of whose entries is $1$. Show that if $n\gt 1$, then
  $$(I-J_n)^{-1}=I-\frac{1}{n-1}J_n$$

Here's what I've tried so far: 
$J_n$ is a matrix of $1$'s, so $I-J_n$ is the matrix whose diagonal entries are $0$ and $1$ elsewhere. But from here, I don't know how to take the inverse of this $n\times n$ matrix. One could use Gaussian-Jordan elimination but I don't think it helps here, it would get too complicated very quickly. Instead, I called the left-hand side of the equation $K$.
\begin{align}
(I-J_n)^{-1} & = K \\
(I-J_n)^{-1}K^{-1} & = KK^{-1} \\
(K(I-J_n))^{-1} & = I \\
(K-KJ_n)^{-1}&=I \\
I & = K-KJ_n
\end{align}
I still need to use the fact that $J_n$ contains only $1$'s but I'm not sure how to do that.
Any help in the right direction is appreciated. 

$J_n^2=nJ_n$, so $(I-J_n)\left(I-\dfrac1{n-1}J_n\right)=I-\dfrac1{n-1}J_n-J_n+\dfrac1{n-1}J_n^2=I$
