How to prove the formula for the Reciprocal Multifactorial constant?

Let's "decipher" the notation of "multifactorial". Assume $k>0$. Any $n>0$ can be written in the form $n=kq+r$ with $q\geqslant 0$ and $1\leqslant r\leqslant k$; then $n\underbrace{!\ldots!}_{k}:=\prod_{j=0}^q(kj+r)$, and $0\underbrace{!\ldots!}_{k}:=1$. But $$\prod_{j=0}^q(kj+r)=k^{q+1}\prod_{j=0}^q\left(j+\frac rk\right)=k^{q+1}\frac{\Gamma(q+1+r/k)}{\Gamma(r/k)}=\frac{k^{q+1}q!}{\mathrm{B}(r/k,q+1)}$$ using the gamma and beta functions. Using the integral representation of the latter, we get

\begin{align*} m(k)&=1+\sum_{r=1}^k\sum_{q=0}^\infty\frac{1}{(kq+r)\underbrace{!\ldots!}_{k}} \\&=1+\sum_{r=1}^k\sum_{q=0}^\infty\frac{1}{k^{q+1}q!}\mathrm{B}\left(\frac rk,q+1\right) \\&=1+\sum_{r=1}^k\sum_{q=0}^\infty\frac{1}{k^{q+1}q!}\int_0^1 t^{r/k-1}(1-t)^q\,dt \\&=1+\frac1k\sum_{r=1}^k\int_0^1 t^{r/k-1}\sum_{q=0}^\infty\frac1{q!}\left(\frac{1-t}{k}\right)^q dt \\&=1+\frac1k\sum_{r=1}^k\int_0^1 t^{r/k-1}e^{(1-t)/k}\,dt\quad\color{gray}{[t=kx]} \\&=1+\frac{e^{1/k}}k\sum_{r=1}^k k^{r/k}\int_0^{1/k}x^{r/k-1}e^{-x}\,dx, \end{align*}

which is the first of the two closed forms. The second one is then easy to obtain: at $r=k$ $$k^{r/k}\int_0^{1/k}x^{r/k-1}e^{-x}\,dx=k\int_0^{1/k}e^{-x}\,dx=k(1-e^{-1/k}).$$