How do I show that these sums are the same? My textbook says that I should check that
$$ \sum_{i=0}^\infty \frac{\left( \lambda\mathtt{I} + \mathtt{J}_k \right)^i}{i!} $$
is in fact the same as the product of sums
$$ \left( \sum_{i=0}^\infty \frac{\left( \lambda\mathtt{I}\right)^i}{i!} \right) \cdot
\left( \sum_{j=0}^k\frac{\left( \mathtt{J}_k \right)^j}{j!} \right)$$
Where $ \mathtt{J}_k $ is all zero exept first super diagonal that has all ones.
But I can't figure out how to do it.
[edit]
To clarify: I'm working towards a definition of $f(\mathtt{A})$ where $f$ is a "nice" function, and $\mathtt{A}$ is an arbitrary square matrix.
The text basically goes like this.
$\mathtt{B} = f(\mathtt{A})$ defined as $b_{ij} = f(a_{ij})$ is a bad idea because 
$f(\mathtt{A})$ where $f(x) = x^2$ is generally not the same as $\mathtt{A}^2$ and so on.
BUT, we know that for numbers, $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} $ so lets try this for matrices.
Then it goes on to show that for diagonal matrices the power series gives the same result as if we would apply the function on the diagonal elements then its expanded to diagonalizable matrices. Then to Jordan blocks, and thats where these sums come in.

Hints:
$$(1)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum_{k=0}^\infty\frac{X^k}{k!}=e^X$$
$$(2)\;\;\;\;\;\;\;\;J_k^{n}=0\;,\;\;\;\text{where$\,n\,$ is the number of rows of the matrix}\;J_k$$
