<s>[INST] Calculating the exponential of an upper triangular matrix 
Find the matrix exponential $e^A$ for
$$ A = \begin{bmatrix}
2 & 1 & 1\\
0 & 2 & 1\\
0 & 0 & 2\\
\end{bmatrix}.$$

I think we should use the proberty

If $AB = BA$ then $e^{A+B} = e^A e^B$.

We can use that
$$\begin{bmatrix}
2 & 1 & 1\\
0 & 2 & 1\\
0 & 0 & 2\\
\end{bmatrix} 
=\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} 
+\begin{bmatrix}
1 & 1 & 1\\
0 & 1 & 1\\
0 & 0 & 1\\
\end{bmatrix}$$
Both matrices obviously commute. But I dont know how to calculate the exponential of
$$\begin{bmatrix}
1 & 1 & 1\\
0 & 1 & 1\\
0 & 0 & 1\\
\end{bmatrix}.$$
Could you help me?
 [/INST] Edit: As pointed out by @XTChen there is a much easier way to do this but I detail evaluation of the requested exponential nonetheless.
Note that
$$\pmatrix{1&1&1\\0&1&1\\0&0&1}^n=\pmatrix{1&n&n(n+1)/2\\0&1&n\\0&0&1}$$
\begin{align}
\sum_{n=0}^\infty & \frac1{n!}\pmatrix{1&1&1\\0&1&1\\0&0&1}^n  \\
&=\sum_{n=0}^\infty\frac1{n!}\pmatrix{1&n&n(n+1)/2\\0&1&n\\0&0&1}\\
&=\pmatrix{1&0&0\\0&1&0\\0&0&1}+\sum_{n=1}^\infty\pmatrix{1/n!&1/(n-1)!&(n+1)/(2(n-1)!)\\0&1/n!&1/(n-1)!\\0&0&1/n!}\\
&=\pmatrix{1&0&0\\0&1&0\\0&0&1}+\pmatrix{e-1&e&3e/2\\0&e-1&e\\0&0&e-1}\\
&=\pmatrix{e&e&3e/2\\0&e&e\\0&0&e}\\
&=\frac12e\pmatrix{2&2&3\\0&2&2\\0&0&2}\\
\end{align}
</s>