Inverse of matrix of ones + nI Having a vector $\mathbf{1} \in \mathbb{R}^{n}$ containing only ones, following equality should be true according to a paper I am currently reading:
\begin{equation}
  \left( nI+\mathbf{1}\mathbf{1}^T \right)^{-1}= \frac{1}{n}\left( I - \frac{1}{2n} \mathbf{1}\mathbf{1}^T \right)
\end{equation}
EDIT: what is the general rule for constructing an inverse of a matrix with $n$ on diagonal and $1$ elsewhere and how is this rule derived?
 A: Notice that the matrix $$P=\frac{11^T}{n}$$ is an orthoprojector, meaning $P^2=P=P^T$.
Such matrices are also called idempotent, and they have just 2 distinct eigenvalues $\{0,1\},\,$ since those are the only 2 idempotent scalars.
You wish to  evaluate the function 
$$f(x)=(n+nx)^{-1}$$
with a matrix argument of $P$, i.e.
$$f(P)=(nI+nP)^{-1}$$
Now the nice thing about an idempotent matrix is that the power series for any function only has 2 terms
$$f(P) = c_0I+c_1P$$ Using Sylvester Interpolation, substitute each eigenvalue into the function and into the power series
$$\eqalign{
 f(1) &= \frac{1}{2n} &= c_0+c_1 \cr
 f(0) &= \frac{1}{n} &=  c_0  \cr
}$$ and solve for the coefficients.
Looks like $c_0$ is already "solved", and the remaining coefficient is simply 
$$c_1 = \frac{1}{2n}-\frac{1}{n} =-\frac{1}{2n}$$
Therefore 
$$\eqalign{
f(P) &= \frac{1}{n}I-\frac{1}{2n}P \cr
  &= \frac{1}{n}\Big(I-\frac{1}{2}P\Big) \cr
  &= \frac{1}{n}\Big(I-\frac{1}{2n}11^T\Big) \cr\cr
}$$
Update
Since you can't find anything on Sylvester interpolation, let me work out a second example. 
Let's say that you wanted to calculate the exponential rather than the inverse of your matrix.
$$\exp(nI+11^T)=\exp(nI+nP)$$
This corresponds to the scalar function
$$f(x)=\exp(n+nx)$$
We have the same 2-term series expansion as last time
$$f(P)=c_0I+c_1P$$
Substituting the eigenvalues yields
$$\eqalign{
 f(1) &= e^{n+1} &= c_0+c_1 \cr
 f(0) &= e^n &= c_0 \cr
}$$
Once again, the $c_0$ is "solved" and the remaining coefficient is
$$c_1=e^{n+1}-e^{n}=e^{n}(e-1)$$
Therefore we have
$$\eqalign{
\exp(nI+11^T)
 &= e^{n}I+e^{n}(e-1)P \cr
 &= e^{n}\Big(I+\frac{e-1}{n}11^T\Big) \cr 
}$$
A: As mentioned in one of the comments, you should consider using the Sherman-Morrison formula or the Woodbury identity which states that for a nonsingular matrix $A$ and column vectors $ b, c$ such that $ {A+bc^\top}$ is nonsingular, 
$$( {A+bc^\top})^{-1}={A^{-1}}-\frac{1}{1+{c^\top A^{-1}b}} {A^{-1}bc^\top A^{-1}}$$
Therefore,
\begin{align}
(n I+\mathbf{11}^\top)^{-1}&=\frac{1}{n} I-\frac{1}{1+\frac{1}{n}{ 1^\top 1}}\frac{1}{n^2}\mathbf{11}^\top
\\&=\frac{1}{n} I-\frac{1}{n+n}\frac{1}{n}\mathbf{11}^\top
\\&=\frac{1}{n}\left( I-\frac{1}{2n}\mathbf{11}^\top\right)
\end{align}

To see how the general formula is derived, first note that 
$$\det (A+bc^\top)\ne 0 \implies 1+{c^\top A^{-1}b}\ne 0$$
Suppose  $A$ is of order $p\times p$, and $b$ and ${c}$ are both $p\times 1$ column vectors.
Let $$d={A+bc^\top}$$
Then,
\begin{align}
{dA^{-1}}&={I_p}+{bc^\top A^{-1}}
\\\\&\implies {dA^{-1}b}=b+{bc^\top A^{-1}b}= b  (1+{c^\top A^{-1}b})
\\\\&\implies ({dA^{-1}b})(1+{c^\top A^{-1}b})^{-1}=b
\\\\&\implies ({dA^{-1}b})(1+{c^\top A^{-1}b})^{-1} c^\top={bc^\top}
\\\\&\implies  A+({dA^{-1}b})(1+{c^\top A^{-1}b})^{-1} c^\top= A+{bc^\top}= d
\\\\&\implies   A= d (1-{A^{-1}b}(1+{c^\top A^{-1}b})^{-1} c^\top)
\\\\&\implies  {I_p}= d (1-{A^{-1}b}(1+{c^\top A^{-1}b})^{-1} c^\top){A^{-1}}
\\\\&\implies  {d^{-1}}=(1-{A^{-1}b}(1+{c^\top A^{-1}b})^{-1} c^\top){A^{-1}}
\end{align}
That is,
\begin{align}
( {A+bc^\top})^{-1}&={A^{-1}}-{A^{-1}b}(1+{c^\top A^{-1}b})^{-1} c^\top{A^{-1}}
\\\\&={A^{-1}}-\dfrac{1}{1+{c^\top A^{-1}b}} {A^{-1}bc^\top A^{-1}}
\end{align}
A: Example
Example for $n=5$:
$$
 \mathbf{A}(n) =
\left(
\begin{array}{ccccc}
 6 & 1 & 1 & 1 & 1 \\
 1 & 6 & 1 & 1 & 1 \\
 1 & 1 & 6 & 1 & 1 \\
 1 & 1 & 1 & 6 & 1 \\
 1 & 1 & 1 & 1 & 6 \\
\end{array}
\right), \qquad
\mathbf{A}^{-1}(n) =
\frac{1}{50}
\left(
\begin{array}{rrrrr}
 9 & -1 & -1 & -1 & -1 \\
 -1 & 9 & -1 & -1 & -1 \\
 -1 & -1 & 9 & -1 & -1 \\
 -1 & -1 & -1 & 9 & -1 \\
 -1 & -1 & -1 & -1 & 9 \\
\end{array}
\right)
$$
Result
In general,
$$
\mathbf{A}_{r,c}(n) =
\begin{cases}
 1 & r\ne c \\
 n+1 & r = c
\end{cases}
\qquad 
\mathbf{A}^{-1}_{r,c}(n) =
\begin{cases}
 -\frac{1}{2n^{2}} & r\ne c \\
 \frac{2n-1}{2n^{2}} & r = c
\end{cases}
$$
Proof strategy: induction
To compute the inverse matrix,
$$
\mathbf{A}^{-1} = \frac{\text{adj } \mathbf{A}} {\det \mathbf{A}}
$$
where the adjugate matrix, $\text{adj } \mathbf{A}$, is the transpose of the matrix of cofactors.


*

*Establish $\det \mathbf{A}(n) = 2n^{-2}$.

*Establish the matrix of minors is an $n\times n$ matrix of the form
$$
\left[ \begin{array}{rrrr}
  \alpha & \beta & -\beta & \dots \\
  \beta & \alpha & \beta & \dots \\
  -\beta & \beta & \alpha \\
  \vdots & \vdots & & \ddots 
\end{array} \right]
$$ 
with
$$
  \alpha = \left(2n-1 \right) n^{n-2}, \qquad \beta = n^{n-2}
$$
