One observation of special type of square matrix exponentiation I was studying the following type of matrices,
$$
A = \begin{pmatrix}
1 & x_{12} & \cdots &x_{1n}\\
0 & x_{22} & \cdots &x_{2n}\\
\vdots\\
0&\cdots&0&x_{nn}
\end{pmatrix},$$
where

*

*$x_{ij} \in [0,1];$

*$x_{jj} \neq 1$ for all $j > 1;$

*$\displaystyle \sum^n_{i=1}x_{ji} = 1$ for all $j \in \{1,2,\cdots,n\}.$
On random simulation, I am getting
$$
\lim_{N \rightarrow \infty} A^N \rightarrow \begin{pmatrix}
1 & 1 & \cdots &1\\
0 & 0 & \cdots &0\\
\vdots\\
0&\cdots&0&0
\end{pmatrix}.
$$
My question is, what is the cause of this behavior?
I tried the following:
Let $A = \mathbf{1} + U$, where $\mathbf{1}$ is the identity matrix of size $n \times n.$
$$
 A^N = (\mathbf{1} + U)^N = \sum^N_{k = 0}\binom{N}{k}U^k
$$ but this is certainly not helping much.
Note that: A can be understood as the State transition matrix of a Markov Chain, where one can stay in its state or go to a lower state with a probability. 
Additional Observation: $\displaystyle \lim_{N\rightarrow \infty} U^N = \mathbf{0}$
Special Thanks: @tsnao for enhancing clarity and pointing errors.
 A: This is not a full answer, but observe that with
$$
A_0 =
\begin{pmatrix}
1 & 0 & \ldots & 0 \\
0 & 0 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & 0
\end{pmatrix}
\quad \text{and} \quad
A_1 =
\begin{pmatrix}
0 & x_{12} & \ldots & x_{1n} \\
0 & x_{22} & \ldots & x_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & x_{nn}
\end{pmatrix}
$$
we have
$$
A = A_0 + A_1.
$$
Note next that $A_1 A_0 = 0$, and therefore (using induction)
$$
A^k = A_0 + A_0 A_1 + A_0 A_1^2 + \ldots + A_0 A_1^{k - 1} + A_1^k.
$$
Since $A_1^k \to 0$, it seems that
$$
\lim_{k \to \infty} A^k = A_0 ( I - A_1 )^{-1}.
$$
That is, if I didn't make some stupid mistake.

Update. After giving it some thought, I have doubts about $A_1^k \to 0$. It is immediate for $2 \times 2$ matrices, but I am not sure whether it holds for other dimensions. Its diagonal surely does converge to zero, but there still are the off-diagonal entries. Maybe I wrote this a little hastily, probably requires some thinking over.
A: (As is also now seen from your answer) I think in your question you actually wanted to impose the condition
$$\sum_{i=1}^j x_{ij} = 1\ \forall j \in \{2,\dots,n\} \tag{1}\label{1} $$
on the column sums of $A$ instead of the condition
$$\sum^n_{i=1}x_{ji} = 1\ \forall j \in \{1,2,\dots,n\} \tag{2}\label{2}$$
on the row sums of $A$.
Condition \eqref{2} makes no sense, already because we do not have $x_{j1}$'s for any $j$. If now the row-sum condition \eqref{2} were replaced by the row-sum condition that the matrix $A$ be stochastic, then $A^N$ would be stochastic for all natural $N$ and hence the limit of $A^N$ as $N\to\infty$ (if this limit exists) would also be a stochastic matrix and thus would differ from your conjectured limit if $n\ge2$.

So, let us assume \eqref{1}. Then your desired conclusion will hold.

Indeed, note that $P:=A^\top=:[p_{ij}]$ is a stochastic matrix, which is the transition matrix of a Markov chain with absorbing state $1$.
Since $p_{ii}<1$ for all $i\ge2$ and $p_{ij}=0$ if $j>i$, we see that for each $i\ge2$ there is some $j<i$ such that $p_{ij}>0$. So, for each $i$ there is some natural number $k$ such that $p^{(k)}_{i1}>0$, where $p^{(k)}_{ij}$ is the $ij$-entry of the matrix $P^k$. So, the singleton set $\{1\}$ is the only closed class. Therefore and because the state space $\{1,\dots,n\}$ of the chain is finite, the only absorbing state $1$ is reached from any state in a finite time with probability $1$ -- see e.g. Section 2.11, p. 102. So, for each $i$ we have $p^{(N)}_{i1}\to1$ as $N\to\infty$ (and hence $p^{(N)}_{ij}\to0$ as $N\to\infty$ for each $j\ge2$).
Thus, your desired conclusion follows.
A: This is to complete the nice answer by tsnao by showing that $A_1^k\to0$ as $k\to\infty$.
To get that conclusion it is enough to assume that the $x_{ij}$'s are any complex numbers such that $$t:=\max_i|x_{ii}|<1. \tag{10}\label{10}$$
Indeed,
$$|A_1|\le tI+|B|, \tag{20}\label{20} $$
where $|X|$ is the matrix of the moduli of the entries of a matrix $X$, $I$ is the identity matrix, $B$ is a matrix with zeros on and below the diagonal (so that $|B|^n=0$), and the matrix comparison in \eqref{20} (and in what follows) is entry-wise.
So, for $k>n$,
$$|A_1^k|\le|A_1|^k\le(tI+|B|)^k
=\sum_{m=0}^k \binom km t^{k-m}|B|^m
=\sum_{m=0}^{n-1} \binom km t^{k-m}|B|^m
\le n k^{n-1} t^{k-(n-1)}|B|^m\to0$$
by \eqref{10} as $k\to\infty$.

A possibly somewhat simpler way to see that $A_1^k\to0$ as $k\to\infty$ is to note that the eigenvalues of the upper-triangular matrix $A_1$ are the $x_{ii}$'s and $0$, and then use the Jordan form of $A_1$ and \eqref{10}.
A: Based on the previous answers by tsnao and myself, one gets another, more elementary proof of your desired conclusion (and actually of a more general statement).
Indeed, by those previous answers,
$$L:=\lim_{k\to\infty}A^k=A_0(I-A_1 )^{-1},$$
where
$$
A_0=
\begin{pmatrix}
1 & 0 & \ldots & 0 \\
0 & 0 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & 0
\end{pmatrix},
$$
so that rows $2,\dots,n$ of the matrix $A_0$ are zero.
So, rows $2,\dots,n$ of the matrix $L$ are zero as well.
But $A^\top$ is a stochastic matrix. So, $L^\top=\lim_{k\to\infty}(A^\top)^k$ is a stochastic matrix, too, and columns $2,\dots,n$ of the matrix $L$ are zero. Thus, we have your desired conclusion.

Looking again back at the previous answer, we see that, for your desired conclusion to hold, it is actually enough that the $x_{ij}$'s be any complex numbers such that
$$\max_i|x_{ii}|<1$$
and
$$\sum_{i=1}^j x_{ij} = 1\ \forall j \in \{2,\dots,n\}.$$
In particular, the nonnegativity of the $x_{ij}$'s is not needed.
A: The answer is quite simple. First observe that $A$ is triangular, hence the spectrum is on the diagonal. From your assumptions, $1$ is a simple eigenvalue and the other eigenvalues belong to $[0,1)$. Decompose
$$A=uv^T+B$$
where $u,v$ are the left/right eigenvectors associated with the eigenvalue $1$ ($Au=u$ and $A^Tv=v$, $v^Tu=1$) and $Bu=0$, $B^Tv=0$. Then
$$A^N=uv^T+B^N$$
where $B^N\to0_n$ as $N\to+\infty$, because $\rho(B)<1$. Hence $A^N\to uv^T$. Conclude by remarking that
$$u=\begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix},\qquad
v=\begin{pmatrix} 1 \\ \vdots \\ 1 \end{pmatrix}.$$
A: I have the following observation. Possibly it could help a bit.
Let $\alpha(n) = (1,1,\cdots,1)$.
If $N \geq 1$, $\alpha(n) A^N = \alpha(n)\times A \times A^{N-1}=\alpha(n)\times A^{N-1}=\cdots= \alpha(n).$
A: Another approach of this problem would be like this. If $A$ and $B$ be two commuting matrices such that $\sum_{i} A_{ij}=1$ and $\sum_{i} B_{ij}=1$ then $\sum_{i} (AB)_{ij}=\sum_{i,k} A_{ik}B_{kj}=\sum_{k} (\sum_{i} A_{ik})B_{kj}=1$. So, the column sum is preserved for all $A^{n}$ (the upper triangular matrix in question).
Now, we can break the matrix as
$A = \begin{pmatrix}
1 & X\\
0 & B\\
\end{pmatrix}$. Where $X=(x_{12},x_{13},...,x_{1n})$ and $B$ is the leftover $(n-1)\times(n-1)$ matrix.
We will show $\lim\limits_{n \to \infty} B^n=0$. Then the 'column sum preserving property' implies that all entries in the first row of $A^n$ tend to $1$.
We have, $\sum_{i} B_{ij}=\alpha_j\leq 1$ as $x_{1j} \geq 0 ,j>1$ (with at least one of $x_{ij}$ being non-zero) . Now, say sum of $j$-th column of $B^{l}, l>1$ is $\beta^{(l)}_j$. So, $\beta^{(l+1)}_j=\sum_{i,k} B_{ik}B^{(l)}_{kj}=\sum_{k}\alpha_{k} B^{(l)}_{kj}$.
As, $\alpha_{k} \leq 1$ (at least one of them is strictly less than 1) ,$\beta^{(l+1)}_j<\sum_{k}B^{(l)}_{kj}=\beta^{(l)}_j$.  Now, $\beta^{(l)}_j-\beta^{(l+1)}_j=\sum_{k} \Delta_kB^{(l)}_{kj} ; \Delta_k=1-\alpha_k\geq 0$ (at least one being non-zero) $;B^{(l)}_{kj}>0$.
This directly implies that $B^{(l)}_{kj} \to 0$ as $l \to \infty$ and so, $B^n \to 0$ as $n \to 0$, which completes the proof.
