7. $N$ !  ends with exactly 2013 zeros, then the maximum value of $\mathbf{N}$ is


【Analysis】First consider 8000!, the number of prime factor 5 is:
$\left[\frac{8000}{5}\right]+\left[\frac{8000}{5^{2}}\right]+\left[\frac{8000}{5^{3}}\right]+\left[\frac{8000}{5^{4}}\right]+\left[\frac{8000}{5^{5}}\right]=1998$, which means 8000! has 1998 zeros at the end, still 15 zeros short.
Consider the number of prime factor $\mathbf{5}$ in 8005 to 8050, among these 10 multiples of 5, except for 8025 and 8050 which have 2 prime factors 5, the rest of the numbers only contain 1 prime factor 5, here we have 12, still 3 short. Thus consider 8055, 8060, 8065, so 8065! first reaches 2013 zeros at the end, thus the maximum value of $\mathrm{N}$ is 8069.