Hmm, using Pari/GP, I just tried to find the coefficients of $ \varphi (t) $
polcoeffs(exp(-x)*sumalt(k=0,exp(-k*x)*x^k/k!*(k+1)^(k-1)*1.0))
(where the polcoeffs-function just retrieves the whole vector of coefficients up to the defined series-precision and I used x instead of t )
The result of this is an (arbitrary near) approximation to the formal powerseries $ \varphi(t) = 1.0 $ So it may be useful to analyze the approximation of that powerseries to the constant in order to evaluate your limit-formula.
[update] If one expands the powerseries for $e^{- k x}$ and collect coefficients at like powers of x then indeed all coefficients at a certain power of x sum to zero, thus vanish (at least heuristically up to $x^{64}$, I did not do final analysis) . Then all derivatives are also zero and even the L'Hospital-rule is of no use here. However, I don't know whether there could be some special reason that this expansion into powerseries might not be feasible/allowed because the result is an infinite sum of formal powerseries. [end update]
[update 2]
$ \varphi(x)=e^{-x}+xe^{-2x}\frac1{1!} +x^2e^{-3x}\frac3{2!} +x^3e^{-4x}\frac{4^2}{3!} + \ldots $
$\begin{array} {rrrr} \varphi(x)= ( & 1 & -x & + \frac{x^2}{2!}&+ \frac{x^3}{3!} & + \ldots )\\ + \frac{x}{1!}(& 1 & -2x & + \frac{2^2x^2}{2!}&+ \frac{2^3x^3}{3!} & + \ldots ) \\ + \frac{3x^2}{2!}(& 1 & -3x & + \frac{3^2x^2}{2!}&+ \frac{3^3x^3}{3!} & + \ldots ) \\ + \ldots \\ \end{array} $
Collecting equal powers of x means to add coefficients along the antidiagonal so
$\begin{array} {rrrr} \varphi(x)= 1 \\ - x (&\frac{1}{1!}& - \frac{1}{1!} ) \\ + x^2 (&\frac{1}{2!}& - \frac{2}{1!}& + \frac{3^1}{2!}) \\ - x^3 (&\frac{1}{3!}& - \frac{2^2}{2!}& + \frac{3^2}{2!}& - \frac{4^2}{3!}) \\ + \ldots \\ \end{array} $
All the parentheses seem to form binomially weighted sums of like powers with alternating signs which are known to evaluate to zero, so the coefficients at each power of x should be zero and if this is so, the taylorseries is then $\varphi(x)=1 $
[end update 2]