Computing $e^{At}$ in terms of matrix Suppose we have a matrix of the form 
$A = I + N$ with $N^2 = 0$.
Compute $e^{At}$ in terms of $N$.
My initial attempt to solve this problem was 
$e^{At} = I + At + \frac{At^2}{2!} + \frac{At^3}{3!} + \cdots$
$\;\;\;\;\; = I + A(1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots) - A$
$\;\;\;\;\; = I - A + Ae^t$
$\;\;\;\;\; = -N + (I + N)e^t$
$\;\;\;\;\; = Ie^t + N(e^t - 1)$
However, the solution that I was provided states that 
by induction, 
$A^k = I + (k-1)N$ 
$e^{At} = e^tI + te^tN$
which confuses me because I thought
$A^k = I + kN$
I am not sure what went wrong with my approach. 
Could someone help me out? 
Thank you for reading.
 A: If $A$ and $B$ commute, then
$$
\begin{align}
e^{A+B}
&=\sum_{n=0}^\infty\frac1{n!}(A+B)^n\\
&=\sum_{n=0}^\infty\frac1{n!}\sum_{k=0}^n\binom{n}{k}A^{n-k}B^k\\
&=\sum_{k=0}^\infty\sum_{n=k}^\infty\frac1{(n-k)!}A^{n-k}\frac1{k!}B^k\\
&=\sum_{k=0}^\infty\sum_{n=0}^\infty\frac1{n!}A^n\frac1{k!}B^k\\[3pt]
&=e^Ae^B
\end{align}
$$
Therefore, since $I$ commutes with every matrix,
$$
\begin{align}
e^{t(I+N)}
&=e^{tI}e^{tN}\\
&=e^tI(I+tN)\\
&=e^t(I+tN)
\end{align}
$$
Or we can
$$
\begin{align}
e^{t(I+N)}
&=\sum_{k=0}^\infty\frac{t^k}{k!}(I+N)^k\\
&=\sum_{k=0}^\infty\frac{t^k}{k!}(I+kN)\\
&=\sum_{k=0}^\infty\frac{t^k}{k!}I+\sum_{k=1}^\infty\frac{t^k}{k!}kN\\
&=\sum_{k=0}^\infty\frac{t^k}{k!}I+\sum_{k=1}^\infty\frac{t^k}{(k-1)!}N\\
&=\sum_{k=0}^\infty\frac{t^k}{k!}I+\sum_{k=0}^\infty\frac{t^{k+1}}{k!}N\\[6pt]
&=e^tI+te^tN
\end{align}
$$
A: You missed the powers of $A$ in the first line. It should be
\begin{eqnarray*}
e^{At} = I + At + \frac{A^{\color{red}{2}}t^2}{2!} + \frac{A^{\color{red}{3}}t^3}{3!} + \cdots.
\end{eqnarray*}
With $A^k=I+kN$ should be quite easy from here ?
