If decimal representation of $2^n$ starts with $7$, what is the first digit in decimal representation of $5^n$?

$ 
\textbf{(A)}\ 1
\qquad\textbf{(B)}\ 3
\qquad\textbf{(C)}\ 5
\qquad\textbf{(D)}\ 7
\qquad\textbf{(E)}\ 9
$
1. We start by noting that the decimal representation of \(2^n\) starts with 7. This implies that there exists some integer \(k\) such that:
   \[
   7 \times 10^k \leq 2^n < 8 \times 10^k
   \]
   Taking the logarithm base 10 of all parts of the inequality, we get:
   \[
   \log_{10}(7 \times 10^k) \leq \log_{10}(2^n) < \log_{10}(8 \times 10^k)
   \]
   Simplifying the logarithms, we have:
   \[
   \log_{10}(7) + k \leq n \log_{10}(2) < \log_{10}(8) + k
   \]
   Since \(\log_{10}(8) = \log_{10}(2^3) = 3 \log_{10}(2)\), the inequality becomes:
   \[
   \log_{10}(7) + k \leq n \log_{10}(2) < 3 \log_{10}(2) + k
   \]

2. Next, we need to determine the first digit of \(5^n\). We use a similar approach by considering the logarithm base 10:
   \[
   \log_{10}(5^n) = n \log_{10}(5)
   \]
   We know that:
   \[
   \log_{10}(2) + \log_{10}(5) = \log_{10}(10) = 1
   \]
   Therefore:
   \[
   n \log_{10}(5) = n (1 - \log_{10}(2))
   \]

3. Substituting the bounds for \(n \log_{10}(2)\) from the previous step:
   \[
   \log_{10}(7) + k \leq n \log_{10}(2) < 3 \log_{10}(2) + k
   \]
   We can rewrite this in terms of \(n \log_{10}(5)\):
   \[
   n (1 - \log_{10}(2)) = n \log_{10}(5)
   \]
   Subtracting the bounds from 1:
   \[
   1 - 3 \log_{10}(2) - k < n \log_{10}(5) \leq 1 - \log_{10}(7) - k
   \]

4. We need to determine the first digit of \(5^n\) based on the above bounds. We know that:
   \[
   1 - 3 \log_{10}(2) \approx 1 - 3 \times 0.3010 = 1 - 0.903 = 0.097
   \]
   and
   \[
   1 - \log_{10}(7) \approx 1 - 0.845 = 0.155
   \]
   Therefore:
   \[
   0.097 + k < n \log_{10}(5) \leq 0.155 + k
   \]

5. Since \(0.097 < n \log_{10}(5) \leq 0.155\), the first digit of \(5^n\) must be 1, as it falls within the range where the first digit is 1.

Conclusion:
\[
\boxed{1}
\]