Induction on $n$ turns out to be a rather inappropriate method for this problem. Why? For one thing because the statement is false for $n=0$, namely $\sum_{k=0}^0\binom0k(-1)^k=\binom00(-1)^0=1\neq0$, so $A(0)$ fails. Not a good start for induction, if you consider $0\in\Bbb N$ (as I do). But as you computed in the question $A(1)$ is true, so somehow you don't need to have $A(n)$ to get $A(n+1)$ when $n=0$. But this does not mean that induction cannot work, as it requires truth of a statement to be hereditary, not falsehood.
Let us try induction, can call $S(n)$ the sums for $n$, so the statement to prove is $S(n)=0$ for all $n\geq1$ (we better forget about the uncooperative $n=0$). Induction calls for Pascal's recurrence, so we compute
$$
\begin{align}
S(n+1)=\sum_{k\geq0}\binom{n+1}k(-1)^k
&=1+\sum_{k\geq1}(\tbinom nk+\tbinom n{k-1})(-1)^k\\
&=\sum_{k\geq0}\tbinom nk(-1)^k+\sum_{l\geq0}\tbinom nl(-1)^{l+1}
=S(n)-S(n).
\end{align}
$$
Now you can invoke the induction hypothesis $S(n)=0$ to show that the resulting value equals$~0$, but frankly there is no pressing need to do so other than the imperative in the title of this question. So we get $S(n+1)=0$ regardless of what $S(n)$ happened to be. This explains the recovery from a false start.
If you must prove this using induction, why not extend the statement to cover other partial sums? Experimenting a bit tells you that the statement to prove ought to be $\sum_{k=0}^m\binom nk(-1)^k=(-1)^m\binom{n-1}m$. The obvious induction to try is on $m$, and for $m=0$ one gets $\binom n0=\binom{n-1}0$ which is OK for all $n$ (even for $n=0$), so this time we are off with a good start. Assuming the result for $m$ one gets
$$
\begin{align}
\sum_{k=0}^{m+1}\tbinom nk(-1)^k
&=(-1)^m\tbinom{n-1}m+(-1)^{m+1}\tbinom n{m+1}\\
&=(-1)^{m+1}(\tbinom n{m+1}-\tbinom{n-1}m)=(-1)^{m+1}\tbinom{n-1}{m+1}
\end{align}
$$
as desired. Since $\binom{n-1}n=0$ for $n>0$, you get your $A(n)$ for $n>0$ as the special case $m=n$.