Finding zeroes for function $f(t) = e^{k(t-1)} -t$ for $k> 0$ analytically

Question

I tried using Lambert W function the following way $$e^{k(t-1)} -t=0$$ $$e^{k(t-1)}=t$$ $$-ke^{-k} = -kte^{-kt}$$ $$W(-ke^{-k}) = W(-kte^{-kt})$$ $$-k = -kt \implies t = 1$$ but this only gives me one trivial root, graphing the function clearly shows that this function has a root $t_0$ where $0\leq t_0< 1$ as well, is there any analytical way I can reach that point? I need a general form of the root in terms of $k$ so I can apply a limit on it, any kind of help would be appreciated.

Answer 1

$$-ke^{-k}=-kte^{-kt}$$

$$-kte^{-kt}=-ke^{-k}$$

$$-kt=W_n(-ke^{-k})\ \ \ \ \ \ \ \ (n\in\mathbb{Z})$$

$$t=-\frac{1}{k}W_n(-ke^{-k})\ \ \ \ \ \ \ \ (n\in\mathbb{Z})$$

The real solutions are $t=-\frac{1}{k}W_0(-ke^{-k})$ and $t=-\frac{1}{k}W_{-1}(-ke^{-k})$.

$t=1$ holds for all $k$. Which of the both solutions is $1$, depends on $k$.