Assume there is a positive integer $n$ s.t. $(I) \;\;$ the leftmost digit in $2^n$ is 5, $(II) \;\;$ the leftmost digit in $5^n$ is 2. Let $k$ and $m$ be the number of digits in $2^n$ and $5^n$ respectively. Therefore $(1) \;\; 5 \cdot 10^{k-1} \leq 2 ^n < 6 \cdot 10^{k-1}$, $(2) \;\; 2 \cdot 10^{m-1} \leq 5^n < 3 \cdot 10^{m-1}$. Multiplying the inequalities (1) and (2), we obtain $(3) \;\; 10^{k+m-1} \leq 10^n < 1,8 \cdot 10^{k+m-1}$, yielding $n = k+m-1$. Hence, according to (1)-(3), we have $10^{k+m-1} \leq 10^n$ only if $5 \cdot 10^{k-1} = 2^n$ and $2 \cdot 10^{m-1} = 5^n$. Both these equations have no solutions in positive integers. Conclusion: There is no positive integer $n$ which satisfies the conditions (I)-(II).