discrete mathematics - Binomial Theorem identities, evaluate the sum

This is a homework problem, please don't blurt out the answer! :)

I've been given the following, and asked to evaluate the sum:

$$\sum_{k = 0}^{n}(-1)^k\binom{n}{k}10^k$$

So, I started out trying to look at this as equivalent to the binomial theorem, in which case, I could attempt something like this: $10^k = y^{n-k}$ but I didn't feel that got me anywhere.

So I started actually evaluating it...

$$(-1)^0\binom{n}{0}10^0 + (-1)^1\binom{n}{1}10^1 + \ldots + (-1)^n\binom{n}{n}10^n$$

So, if I'm thinking correctly, all the other terms cancel out and you are left with:

$$(-1)^n\binom{n}{n}10^n = (-1)^n10^n$$

But, obviously this cannot be correct (or can it?). The book gives a slightly different answer, so I'm wondering where I'm going wrong. Some direction would be greatly appreciated!

Books answer: $\displaystyle (-1)^n9^n$

Answer

Try to fit your sum into one of the following:
$$

\sum_{k=0}^n\binom{n}ka^kb^{n-k}=(a+b)^n,\quad\sum_{k=0}^n\binom{n}kb^{n-k}=(1+b)^n,\quad\sum_{k=0}^n\binom{n}ka^k=(a+1)^n.
$$