Solving an equation involving LambertW Function. Below i have an equation where $H$ and $k$ is a positive real integers and the $t$ is the changing variable.
$$t^{k-1} e^{-t} = H$$
So when I solve this on Maple in order to isolate for $t$ i get the result presented below:
$$t = -(k-1)LambertW\Bigg(-\frac{e^{-\frac{ln H}{k-1}}}{k-1}\Bigg)$$
My question is how does Maple arrive to this solution? What is confusing me is the $LambertW$ function, what is the condition such that i obtain the following solution? 
 A: $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
Not sure, how exactly the Maple algorithms work,
but typical transformations of the given equation 
to the form $u\,\e^u=v$ 
in order to apply the Lambert $\W$ function
looks something like this:
\begin{align} 
t^{k-1} e^{-t} &= H
,\\
t\exp\left(-\frac t{k-1}\right) &= H^{\frac 1{k-1}}
\quad\text{note that k=1 is a special simpler case}
,\\
-\frac t{k-1}\,\exp\left(-\frac t{k-1}\right) 
&= -\frac 1{k-1}\,H^{\frac 1{k-1}}
\end{align}
At this point we have the desired form $u\,\e^u=v$ of the original equation,
where 
\begin{align} 
u&=-\frac t{k-1}
,\\
v&=-\frac 1{k-1}\,H^{\frac 1{k-1}}
\end{align}
and we can apply the Lambert $\W$ function to get $u$ on the left as
\begin{align} 
\W(u\,\e^u)&=\W(v)
,\\
u&=\W(v)
,
\end{align}
so
\begin{align} 
\W\left(-\frac t{k-1}\,\exp\left(-\frac t{k-1}\right)\right) 
&= \W\left(-\frac 1{k-1}\,H^{\frac 1{k-1}}\right)
,\\
-\frac t{k-1} &= \W\left(-\frac 1{k-1}\,H^{\frac 1{k-1}}\right)
,\\
t &= (1-k)\,\W\left(-\frac 1{k-1}\,H^{\frac 1{k-1}}\right)
.
\end{align} 
At tis point we have the solution of the original equation 
in terms the Lambert $\W$ function 
and we need to make one more important step:
it's time to analyze the argument $v$ of $\W$ to find out 
the number of the real solutions.
It is well-known that 
\begin{align}
v<-\frac1\e\quad&\Rightarrow\quad\text{no real solutions}
,\\
v\ge0\quad&\Rightarrow\quad\text{one real solution, }\Wp(v) \text{ or just } \W(v)
,\\
v\in(-\tfrac1\e,0)
\quad&\Rightarrow\quad\text{two real solutions, }\Wp(v)\in(-1,0) \text{ and } \Wm(v)<-1
,\\  
v=-\tfrac1\e
\quad&\Rightarrow\quad\text{a special case, one real solution, }
\Wp(-\tfrac1\e)=\Wm(-\tfrac1\e)=-1
.
\end{align} 
$\endgroup$
