Similarity between $I+N$ and $e^N$ when $N$ is nilpotent Let
$$
N=\begin{pmatrix}0&1&&\\&\ddots&\ddots&\\&&0&1\\&&&0
\end{pmatrix}_{n\times n}
$$
and $I$ is the identity matrix of order $n$. How to prove $I+N\sim e^N$?
Clarification: this is the definition of similarity, which is not the same as equivalence.
Update: 
I noticed a stronger relation, that $A\sim N$, if
$$
A=\begin{pmatrix}0&1&*&*\\&\ddots&\ddots&*\\&&0&1\\&&&0
\end{pmatrix}_{n\times n}
$$
and $*$'s are arbitrary numbers.

By subtracting $I$ this is equivalent to asking about the similarity class of a nilpotent square matrix of size $N$.    The similarity type of $N$ is determined by the dimensions of the kernels of powers of $N$.  In the upper triangular case the list of dimensions of $\ker N^i$ is $1,2,3,4,...,n$ for both of the matrices you consider.  Hence they are similar. 
