1. The problem is to understand and express the function $\cosh^n(x)$, which means the hyperbolic cosine of $x$ raised to the power $n$. 2. Recall the definition of hyperbolic cosine: $$\cosh(x) = \frac{e^x + e^{-x}}{2}$$ 3. Therefore, $$\cosh^n(x) = \left(\frac{e^x + e^{-x}}{2}\right)^n$$ 4. This expression can be expanded using the binomial theorem if needed: $$\cosh^n(x) = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} e^{x(n-2k)}$$ 5. This formula shows that $\cosh^n(x)$ is a sum of exponential functions with coefficients given by binomial coefficients. 6. This is useful in many areas such as solving differential equations or evaluating integrals involving powers of hyperbolic cosine. Final answer: $$\cosh^n(x) = \left(\frac{e^x + e^{-x}}{2}\right)^n = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} e^{x(n-2k)}$$