$\newcommand{\kinv}{(\frac{1}{k})}$ Take the equation $S_k = \frac{k-1}{k!}(\kinv^0+\kinv^1+\kinv^2+\cdots)$. Evaluate it for $k=1$, and we can see that $k-1=1-1=0$, so the whole thing must equal $0$.
Good. Now solve the geometric series and we find that $S_k = \frac{k-1}{k!}(\frac{k}{k-1})$. But if we evaluate this at $k=1$, then $S_k$ is undefined because we have $0$ in the denominator!
Okay, so what to do? Let's simplify to this: $S_k = \frac{k}{k!}$. So now we can evaluate it once again, and it is not undefined, but rather $1/1!=1$. Wait, what? Our previous, equivalent formula gave us $0$, not $1$!
Okay, what if we cancel the k's? Then we'd have $S_k = \frac{k}{k(k-1)(k-2)\cdots} = \frac{1}{(k-1)!}$. And here, once again, we have a $0$ in the denominator if we try to evaluate, and so the solution is once again undefined. :-(
Can someone explain why these equations don't always evaluate the same, even though they are valid and equivalent?
Correction: As others have pointed out, the 4th iteration of the formula actually evaluates to $1$.