Trying to solve a demonstration with the constant of Euler I think this will be an easy problem for you, but I do not see the solution.
I know that
$$\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}=e$$
Knowing this, how can I demonstrate this
$$\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}\frac{1}{k!}=1-\frac{1}{e}$$
this was my attempt 
i) $$\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}=e$$
ii)$$\displaystyle\sum_{k=1}^{\infty} \frac{1}{k!}=1+e$$
iii)$$\Bigg[\displaystyle\sum_{k=1}^{\infty}\frac{1}{k!}\Bigg]^{-1}=\frac{1}{1+e}$$
And I'm stuck here. I certainly took the wrong path. Can you help me?

Hint. Starting with
$$\sum_{k=1}^{\infty} (-1)^{k-1}\frac{1}{k!}=1-\frac{1}{e}$$
Move the $1$ to the left:
$$\sum_{k=0}^{\infty} (-1)^{k-1}\frac{1}{k!}=-\frac{1}{e}$$
Get rid of the minus sign:
$$\sum_{k=0}^{\infty} (-1)^k\frac{1}{k!}=\frac{1}{e}$$
Now, multiply the series for $e$ and $\frac{1}{e}$, the result should end up being $1$:
$$e\cdot\frac{1}{e} = \left(\sum_{k=0}^{\infty}\frac{1}{k!}\right)\cdot\left(\sum_{k=0}^{\infty} (-1)^k\frac{1}{k!}\right)\overset{?}=\dots\overset{?}=1$$
