Let $a_m(n,k)$ be the number of set partitions of $n$ elements into $k$ non-empty partitions with maximum size $m$ each. We show the following is valid for $m\geq 1$:
\begin{align*}
\color{blue}{a_m(n+1,k)}&\color{blue}{=ka_m(n,k)+a(n,k-1)-\binom{n}{m}a(n-m,k-1)\qquad n,k\geq 1}\tag{1}
\end{align*}
with boundary conditions
\begin{align*}
\color{blue}{a_m(n,k)}&\color{blue}{=a_m(n,k)=0\, \qquad\qquad n<k,n>km}\\
\color{blue}{a_m(n,n)}&\color{blue}{=1\, \qquad\qquad\qquad\qquad\quad n\geq 0}\\
\color{blue}{a_m(n,1)}&\color{blue}{=}
\color{blue}{\begin{cases}
1&\qquad\qquad\qquad\quad 1\leq n\leq m\\
0&\qquad\qquad\qquad\quad n>m
\end{cases}}\\
\color{blue}{a_m(0,k)}&\color{blue}{=a_m(n,0)=0\qquad\qquad\ \, n,k>0}\\
\end{align*}
This approach is based upon generating functions. The following can be found in section II.3.1 in Analytic combinatorics by P. Flajolet and R. Sedgewick:
The class $S^{(A,B)}$ of set partitions with block sizes in $A\subseteq \mathbb{Z}_{\geq 1}$ and with a number of blocks that belongs to $B$ has exponential generating function
\begin{align*}
S^{(A,B)}(z)=\beta(\alpha(z))\qquad\text{where}\qquad \alpha(z)=\sum_{a\in A}\frac{z^a}{a!},\quad \beta(z)=\sum_{b\in B}\frac{z^b}{b!}\tag{2}
\end{align*}
At first we determine a generating function for $a_m(n,k)$.
Generating function: In the current situation we are looking for partitions with maximum size $m$. We set
\begin{align*}
A=\{1,2,\ldots,m\}\qquad\text{where}\qquad\alpha(z)=\sum_{j=1}^m\frac{z^j}{j!}
\end{align*}
accordingly. Since the number of partitions is $k$ we have
\begin{align*}
B=\{k\}\qquad\text{where}\qquad \beta(z)=\frac{z^k}{k!}
\end{align*}
The resulting generating function is according to (2)
\begin{align*}
\beta(\alpha(z))=\frac{1}{k!}\left(\sum_{j=1}^m\frac{z^j}{j!}\right)^k=\sum_{n=k}^{km}a_m(n,k)\frac{z^n}{n!}\tag{3}
\end{align*}
We can use the generating function (3) to obtain a recurrence relation for $a_m(n,k)$.
Recurrence relation:
We obtain
\begin{align*}
\frac{d}{dz}\sum_{n=k}^{km}a_m(n,k)\frac{z^n}{n!}&=\sum_{n=k}^{km}a_m(n,k)\frac{z^{n-1}}{(n-1)!}
=\sum_{n=k-1}^{km-1}a_m(n+1,k)\frac{z^n}{n!}\tag{4}
\end{align*}
On the other hand we obtain
\begin{align*}
\frac{d}{dz}&\left(\frac{1}{k!}\left(\sum_{j=1}^m\frac{z^j}{j!}\right)^k\right)\\
&=\frac{1}{(k-1)!}\left(\sum_{j=1}^m\frac{z^j}{j!}\right)^{k-1}\sum_{j=1}^m\frac{z^{j-1}}{(j-1)!}\\
&=\frac{1}{(k-1)!}\left(\sum_{j=1}^m\frac{z^j}{j!}\right)^{k-1}\sum_{j=0}^m\frac{z^{j}}{j!}\\
&=\frac{1}{(k-1)!}\left(\sum_{j=1}^m\frac{z^j}{j!}\right)^{k-1}\left(\sum_{j=1}^m\frac{z^{j}}{j!}+1-\frac{z^m}{m!}\right)\\
&=\frac{1}{(k-1)!}\left(\sum_{j=1}^m\frac{z^j}{j!}\right)^k+\frac{1}{(k-1)!}\left(\sum_{j=1}^m\frac{z^j}{j!}\right)^{k-1}\\
&\qquad-\frac{z^m}{m!}\cdot\frac{1}{(k-1)!}\left(\sum_{j=1}^m\frac{z^j}{j!}\right)^{k-1}\\
&=k\sum_{n=k}^{km}a_m(n,k)\frac{z^n}{n!}+\sum_{n=k-1}^{(k-1)m}a_m(n,k-1)\frac{z^n}{n!}
-\frac{z^m}{m!}\sum_{n=k-1}^{(k-1)m}a_m(n,k-1)\frac{z^n}{n!}\tag{5}
\end{align*}
The right-most series in (5) is
\begin{align*}
\frac{1}{m!}\sum_{n=k-1}^{(k-1)m}a_m(n,k-1)\frac{z^{n+m}}{n!}
&=\frac{1}{m!}\sum_{n=k+m-1}^{km}a_m(n-m,k-1)\frac{z^{n}}{(n-m)!}\\
&=\binom{n}{m}\sum_{n=k+m-1}^{km}a_m(n-m,k-1)\frac{z^{n}}{n!}\\
\end{align*}
We conclude from (4) and (5)
\begin{align*}
\sum_{n=k-1}^{km-1}a_m(n+1,k)\frac{z^n}{n!}
&=k\sum_{n=k}^{km}a_m(n,k)\frac{z^n}{n!}+\sum_{n=k-1}^{(k-1)m}a_m(n,k-1)\frac{z^n}{n!}\\
&\qquad -\binom{n}{m}\sum_{n=k+m-1}^{km}a_m(n-m,k-1)\frac{z^{n}}{n!}\tag{6}
\end{align*}
Coefficient comparison of (6) for $k\leq n\leq km$ results in
\begin{align*}
\color{blue}{a_m(n+1,k)=ka_m(n,k)+a_m(n,k-1)-\binom{n}{m}a_m(n-m,k-1)}
\end{align*}
and the claim follows when also respecting the boundary conditions stated in (1).
Best Answer
Using the notation from Analytic Combinatorics we get the combinatorial class
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}_{=k}(\textsc{SET}_{\text{even}}(\mathcal{Z})).$$
Translating to generating functions we then obtain
$$\frac{1}{k!} \left(\sum_{q\ge 1} \frac{z^{2q}}{(2q)!}\right)^k \\ = \frac{1}{k!} \left(-1 + \sum_{q\ge 0} \frac{z^{2q}}{(2q)!}\right)^k \\ = \frac{1}{k!} \left(-1 + \frac{1}{2}(\exp(z)+\exp(-z))\right)^k \\ = \frac{1}{k!} \left(-1 + \cosh(z)\right)^k.$$
Odd is derived similarly.