Split $1000 = 8 \cdot 125:$ for $n \ge 3$ you need $n \equiv 5^n \pmod 8$ and $n \equiv 2^n \pmod{125}$

$2^n \bmod 125$ depends only on $n \bmod 100.$ Force consistency modulo $5,$ then $25,$ to pin down $n \bmod 100,$ and finish with CRT.

Work modulo $8$ and $125.$ For $n \ge 3$ we have $2^n \equiv 0 \pmod 8,$ so we need $n \equiv 5^n \pmod 8.$ If $n$ is even then $5^n \equiv 1,$ forcing the even number $n$ to be $\equiv 1 \pmod 8,$ impossible; so $n$ is odd, $5^n \equiv 5,$ and $n \equiv 5 \pmod 8.$ Also $5^n \equiv 0 \pmod{125}$ for $n \ge 3,$ so we need $n \equiv 2^n \pmod{125}.$

The order of $2$ is $4$ modulo $5,$ $20$ modulo $25,$ and $100$ modulo $125.$ Since $n \equiv 5 \pmod 8$ gives $n \equiv 1 \pmod 4,$ we get $2^n \equiv 2 \pmod 5,$ so $n \equiv 2 \pmod 5$ and hence $n \equiv 17 \pmod{20}.$ Then $2^n \equiv 2^{17} = 2^{10} \cdot 2^7$ $\equiv (-1)(3) \equiv 22 \pmod{25},$ so $n \equiv 22 \pmod{25},$ which with $n \equiv 1 \pmod 4$ gives $n \equiv 97 \pmod{100}.$ Finally $2^{10} \equiv 24,$ $2^{20} \equiv 76,$ $2^{40} \equiv 26,$ $2^{80} \equiv 51 \pmod{125},$ so $$\begin{aligned} &2^n \equiv 2^{97} = 2^{80} \cdot 2^{10} \cdot 2^7 \\ &\equiv 51 \cdot 24 \cdot 3 \equiv 47 \pmod{125}, \end{aligned}$$ giving $n \equiv 47 \pmod{125}.$

Combining $n \equiv 47 \pmod{125}$ with $n \equiv 5 \pmod 8$ yields $n \equiv 797 \pmod{1000},$ and $n = 1, 2$ fail by direct check, so the least such $n$ is $797.$