Inverse of an upper triangular matrix with all entries 1 I want to find the inverse of a large upper triangular matrix where all its entries are 1.
Is there some trick to it or do I have to compute it using the usual way?

If you look at matrix multiplication as explained in this post, you can pretty much see what the inverse in this case. I will give a detailed explanation here, but I would normally go through this in my head.
Let A be the matrix you described, in the case of $n=3$, then we have
$\newcommand{\vek}[1]{\boldsymbol{#1}}$
\begin{align}
A =
\begin{pmatrix}
 &  &   \\
\vek{a}_1 & \vek{a}_2 & \vek{a}_3 \\
 &  &   \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 1 & 1 \\
0 & 1 &  1 \\
0 & 0 & 1 \\
\end{pmatrix}
\end{align}
Also let $\vek{e}_j$ denote the canonical base vectors of the three dimensional vector space. If you red the post I linked, you should realise that, calculating the inverse comes down to finding $\vek{\nu}_j$'s such that 
$$
A 
\begin{pmatrix}
 &  &   \\
\vek{\nu}_1 & \vek{\nu}_2 & \vek{\nu}_3 \\
 &  &   \\
\end{pmatrix}
=
\begin{pmatrix}
 &  &   \\
A\vek{\nu}_1 & A\vek{\nu}_2 & A\vek{\nu}_3 \\
 &  &   \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
$$
Because then the inverse of $A$ is just the matrix that has the $\vek{\nu}_j$ as column vectors. Since every $\vek{\nu_j}$ gives us a linear combinations of the $\vek{a}_k$'s, we will just try to find linear combinations of the $\vek{a}_k$'s that are equal to $\vek{e}_j$ for $j= 1,2,3$
\begin{align}
\vek{e}_1 
=
\begin{pmatrix}
1 \\ 0 \\ 0
\end{pmatrix}
&=
\vek{a}_1 = A
\begin{pmatrix}
1 \\ 0 \\ 0
\end{pmatrix}
\\
\vek{e}_2 
= 
\begin{pmatrix}
1 \\ 1 \\ 0
\end{pmatrix}
-
\begin{pmatrix}
1 \\ 0 \\ 0
\end{pmatrix}
&=
\vek{a}_2 - \vek{a}_1 
= A
\begin{pmatrix}
-1 \\ 1 \\ 0
\end{pmatrix}
\\
\vek{e}_3 
=
\begin{pmatrix}
1 \\ 1 \\ 1
\end{pmatrix}
-
\begin{pmatrix}
1 \\ 1 \\ 0
\end{pmatrix}
&= 
\vek{a}_3 - \vek{a}_2 
= A
\begin{pmatrix}
0 \\ -1 \\ 1
\end{pmatrix}
\end{align}
So the inverse turns out to be
$$
\begin{pmatrix}
1 & -1 & 0 \\
0 & 1 &  -1 \\
0 & 0 & 1 \\
\end{pmatrix}
$$
Just as a check, using the calculations we did above, we get
$$
A \begin{pmatrix}
1 & -1 & 0 \\
0 & 1 &  -1 \\
0 & 0 & 1 \\
\end{pmatrix}
=
\begin{pmatrix}
A
\begin{pmatrix}
1 \\ 0 \\ 0
\end{pmatrix} &
A
\begin{pmatrix}
-1 \\ 1 \\ 0
\end{pmatrix} &
A
\begin{pmatrix}
0 \\ -1 \\ 1
\end{pmatrix}
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
$$
For $n>3$ you should now be able to see how to get the $\vek{e}_j$'s, using the $\vek{a}_k$'s.
