For a constant $c$, a sequence $a_n$ is defined by $a_n=\int_c^1 nx^{n-1}\left(\ln \left(\frac{1}{x}\right)\right)^n dx\ (n=1,\ 2,\ 3,\ \cdots).$

Find $\lim_{n\to\infty} a_n$.
1. Consider the given sequence \( a_n = \int_c^1 nx^{n-1} \left( \ln \left( \frac{1}{x} \right) \right)^n \, dx \).

2. To simplify the integral, we perform a substitution. Let \( u = \ln \left( \frac{1}{x} \right) \). Then \( du = -\frac{1}{x} dx \) or \( dx = -e^{-u} du \). Also, when \( x = c \), \( u = \ln \left( \frac{1}{c} \right) \), and when \( x = 1 \), \( u = 0 \).

3. Substituting these into the integral, we get:
   \[
   a_n = \int_{\ln \left( \frac{1}{c} \right)}^0 n e^{-u(n-1)} u^n (-e^{-u}) \, du
   \]
   Simplifying, we have:
   \[
   a_n = n \int_{\ln \left( \frac{1}{c} \right)}^0 e^{-un} u^n \, du
   \]

4. Change the limits of integration to make the integral more manageable:
   \[
   a_n = n \int_0^{\ln \left( \frac{1}{c} \right)} e^{-un} u^n \, du
   \]

5. To evaluate the limit as \( n \to \infty \), we analyze the behavior of the integrand \( e^{-un} u^n \). For large \( n \), the term \( e^{-un} \) decays very rapidly unless \( u \) is very small. Therefore, the main contribution to the integral comes from the region where \( u \) is close to 0.

6. Consider the behavior of the integrand near \( u = 0 \):
   \[
   e^{-un} u^n \approx u^n \quad \text{for small } u
   \]

7. As \( n \to \infty \), the term \( u^n \) becomes very small for any fixed \( u \) in the interval \( [0, \ln \left( \frac{1}{c} \right)] \). Hence, the integral \( \int_0^{\ln \left( \frac{1}{c} \right)} e^{-un} u^n \, du \) tends to 0.

8. Therefore, we conclude that:
   \[
   \lim_{n \to \infty} a_n = \lim_{n \to \infty} n \int_0^{\ln \left( \frac{1}{c} \right)} e^{-un} u^n \, du = 0
   \]

The final answer is \(\boxed{0}\)