Show that exponential map between complex matrix and general linear group is surjective $\textbf{Problem}$
Trying to that $\exp: M_n(\mathbb{C}) \rightarrow GL(n,\mathbb{C})$  is surjective.
$\textbf{Attempt}$
I started by considering a complex matrix A $\in M_n \mathbb{C}$, to this matrix I applied the Jordan normal form so $ A = \lambda I + N $ where $I$ is the identity matrix and $N$ is an upper triangular matrix.
Thus, using the exponential map I get:
$$ M_n \mathbb{C} \rightarrow GL(n \mathbb{C}) \\
A = (\lambda I + N) \mapsto exp(\lambda I + N)
$$
Expanding the exponential we get and using the following Jordan Formula:
$\textbf{Jordan Formula:}$
$$\mathbf J=\begin{pmatrix}\lambda&1&&\\&\lambda&\ddots&\\&&\ddots&1\\&&&\lambda\end{pmatrix}$$
$$f(\mathbf A)=\begin{pmatrix}f(\lambda)&f^\prime(\lambda)&\cdots&\frac{f^{(n-1)}(\lambda)}{(n-1)!}\\&f(\lambda)&\ddots&\vdots\\&&\ddots&f^\prime(\lambda)\\&&&f(\lambda)\end{pmatrix}$$
$$exp(\mathbf A)=\begin{pmatrix}exp(\lambda)&exp(\lambda)&\cdots&\frac{exp(\lambda)}{(n-1)!}\\&exp(\lambda)&\ddots&\vdots\\&&\ddots&exp(\lambda)\\&&&exp(\lambda)\end{pmatrix}$$
and I can identify this last matrix as an element of $GL(N, \mathbb C)$? If yes, is this enough to prove the surjectiveness of the map? What am I missing?
Thanks in advance!
Edit:
I used @BenGrossmann and @user1551 hint, so I defined:
$ M = log(\lambda I ) + N $ and $J = \lambda I + N $ then
$\exp(M) = \exp(\log ( \lambda I) + N) = \lambda I \exp(N) $ as N is nilpotent of degree 3 thus $exp(N) \approx I + N + \frac{1}{2} N^2$, I plug this in the above equation and I get:
$$ \exp(M) = \lambda I (\approx I + N + \frac{1}{2} N^2) $$
How can I prove that $\exp(M) = J$ from this?
 A: Let $A$ be any  unital, commutative algebra over any field with characteristic zero.  Denote by
$$
  N = \{x\in A: x^n=0,  \text{ for some } n>0\}
  $$
and
$$
  U = \{y\in A: (y-1)^n=0,  \text{ for some } n>0\}.
  $$
Since $A$  commutative, it easy easy to see that $N$ is a subalgebra (the fact that it is closed under addition follows
from  Newton's Binomial Theorem).
Observe also  that  $U=1+N$.
Next consider the functions
$$
  \ln:U\to N, \quad\text{and}\quad
  \exp:N\to U,
  $$
given by
$$
  \ln(y) = -\sum_{n=1}^\infty \frac{(1-y)^n}n, \quad\text{and}\quad
  \exp(x) = \sum_{n=0}^\infty   \frac{x^n}{n!}.
  $$
Despite the occurence of "$\infty $" as the upper limits of the above sums, only finitely many terms are nonzero because
$1-y$ and $x$ are  nilpotent elements.  Notice that
the fact that $N$ is an ideal guarantees that  $\ln$ is indeed  $N$-valued, while  $\exp$ is $U$-valued.
Although it is a bit of a pain to prove it, these maps are each others inverse.
Given an invertible complex matrix $y$ whose Jordan form consists of a single Jordan block,  we may write
$$
  y=\lambda I+x
  $$
where  $\lambda \neq 0$ and $x$ is nilpotent.  Rewritting $y$ as
$$
  y=\lambda (I+x/\lambda ),
  $$
writing $\lambda =\exp(z)$, for some complex number $z$, and  working within the unital commutative  algebra of matrices generated by $y$, we then have that
$$
  y=\lambda (I+x/\lambda ) = $$$$ =
  \exp(z) \exp\big (\ln(I+x/\lambda )\big ) = $$$$ =
  \exp\big (z+ \ln(I+x/\lambda )\big ).
  $$
For a general invertible matrix $y$, one may repeat the above procedure for each of its Jordan blocks.
