Let $a_n=\frac{{1000}^n}{n!}$ for $n\in N$ , then $a_n$ is greatest when n is/are

$\text{ For }{a}_n\text{ to be greatest, consider this }{a}_n\ge a_{n-1}\text{ i.e }\left(a_n\text{ increases with increase in }n\right.\text{ ) }$

$\frac{{1000}^{\mathrm{n}}}{\left(\mathrm{n}\right)!}\ge \frac{{1000}^{\mathrm{n}-1}}{\left(\mathrm{n}-1\right)!}$

$\begin{array}{r}\frac{1000}{n}\ge 1\\ 1000\ge \mathrm{n}\\ \mathrm{n}\le 1000\end{array}$

$\frac{{1000}^{1000}}{1000!}=\left(\frac{1000}{1}\right)\cdot \left(\frac{1000}{2}\right)\cdot \left(\frac{1000}{3}\right)\cdot 0\cdot \left(\frac{998}{1000}\right)\cdot \left(\frac{999}{1000}\right)\cdot \left(\frac{1000}{1000}\right)$

$\text{ which is equal to the 999th term since the last factor is }\frac{1000}{1000}\text{ is }1\text{. }$

$$\begin{gathered} \text{ After the 1000th term we will be multinlving by a fraction less than }1\text{. } \\ \text{ beginning with multiplying by }\frac{1000}{1001}\text{ and so the sequence will } \\ \text{ begin to decrease with the 1001st term. } \end{gathered}$$

$$\begin{gathered} \text{ So the }999\text{ th and the }1000\text{ th terms, which are equal, are the largest} \\ \text{ terms of the sequence. Theretore the sequence attains maximum at } \\ \mathrm{exactly} \text{two values of }n,999\text{ and }1000\text{. } \end{gathered}$$