linear algebra - condition number of a matrix with diagonal ones and constant else.

Consider the matrix
$$A=\left(\begin{array}{ccccc}
1 & c & \cdots & c\\
c & 1 & \ddots & \vdots \\

\vdots & \ddots & \ddots & c\\
c & \cdots & c & 1\\
\end{array}\right)$$
for some $c\geq1$. So $A$ is a matrix with every entry equal to $c$, excepts for the diagonal entries, who equal $1$.

If $c=1$, $A$ is singular. Furthermore if $c$ increases, the condition number of $A$ decreases. I want to prove this, but I do not know how. Manually writing out the condition number seems a burden.

So my question is: how do I compute the condition number of $A$?

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Edit:
The solution is
$$\kappa(A)=\left|\frac{\lambda_{\mathrm{max}}(A)}{\lambda_{\mathrm{min}}(A)}\right|=\left|\frac{(N-1)c+1}{1-c}\right|=\frac{Nc}{c-1}-1,$$
since the eigenvalues of $A$ are $1-c$ and $(N-1)c+1$, where $N$ is the number of rows (or columns) of $A$.

However, I retrieved the eigenvalues by throwing a couple of matrices into WolframAlpha. So my question reduces to how to prove $1-c$ and $(N-1)c+1$ are the eigenvalues of $A$?