How many zeros are there at the end of the number $9^{2007}+1$?
Initially, we verify how the powers of 9 end, that is, we list the last two digits, the tens and units, of the powers $9^{n}$, in order.

| If $n$ is | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $9^{n}$ ends with | 01 | 09 | 81 | 29 | 61 | 49 | 41 | 69 | 21 | 89 | 01 | 09 | 81 |

Thus, we see that the last two digits of $9^{10}, 9^{11}$, and $9^{12}$ are the same as those of $9^{0}, 9^{1}$, and $9^{2}$. Starting from $9^{10}$, the last two digits of the powers begin to repeat, forming a periodic sequence with a period of 10. Since $2007=10 \times 200+7$ and the last two digits of $9^{10 \times 200}$ are 01, it follows that the last two digits of $9^{2007}$ are the last two digits of $9^{7}$, which are 69. Therefore, the last two digits of $9^{2007}+1$ are equal to $69+1=70$. Thus, there is a single zero at the end of the number $9^{2007}+1$.