Similarity between $I+N$ and $e^N$ when $N$ is nilpotent

Question

Let $$ N=\begin{pmatrix}0&1&&\\&\ddots&\ddots&\\&&0&1\\&&&0 \end{pmatrix}_{n\times n} $$ and $I$ is the identity matrix of order $n$, can anyone show me how to prove $I+N\sim e^N$? Thank you~

Further, can this be generalized to exponential mapping or something like this?

Clarification: Terms may differ from place to place, but this is the definition of similarity here, which is not the same as equivalence. Those who wants to verify the similarity can use the following Mathematica snippet

NilpotentMatrix[n_] := Table[Boole[j == i + 1], {i, n}, {j, n}]; JordanDecomposition[MatrixExp[NilpotentMatrix[#]]][[2]]& [AnyOrderYouWant] //MatrixForm

to see whether the Jordan canonical form of $e^N$ is equal to $I+N$.

Update 2: $I$ subtracted.

Update: By using

JordanDecomposition[Table[Boole[i <= j <= i + 1] +10 Boole[j >= i + 2] (RandomReal[] - 0.5), {i, #}, {j, #}]][[2]] &[AnyOrderYouWant] //MatrixForm

I noticed a stronger proposition, that $A\sim N$, where $$ A=\begin{pmatrix}0&1&*&*\\&\ddots&\ddots&*\\&&0&1\\&&&0 \end{pmatrix}_{n\times n} $$ and $*$'s stand for an arbitrary triangle.

Answer 1

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.