Determinant of a $4 \times 4$ matrix $A$ and $(\det(A))^5$ Calculate $\det(A)$ and $\det(A)^5$:
$$A= \begin{bmatrix}a&a&a&a\\a&b&b&b\\a&b&c&c\\a&b&c&d\end{bmatrix}$$
I found $\det A$ with Laplace expansion: $$a(-abc+b^2c+a c^2-b c^2+a b d-b^2 d-a c d+b c d) .$$
But how can I determine $\det A^5$ easily/fast? I know that there is the possibility to use $\det(A^5)=\det(A)^5)$, but this is too long for given time to resolve the problem.

You can much easier get your expression by subtracting the third line from the fourth, then the second from the third and finally, the first from the second:
$$
\det A=
\det \begin{pmatrix}
a&a&a&a\\
a&b&b&b\\
a&b&c&c\\
0&0&0&d-c
\end{pmatrix}=
\det \begin{pmatrix}
a&a&a&a\\
a&b&b&b\\
0&0&c-b&c-b\\
0&0&0&d-c
\end{pmatrix}=
\det \begin{pmatrix}
a&a&a&a\\
0&b-a&b-a&b-a\\
0&0&c-b&c-b\\
0&0&0&d-c
\end{pmatrix}=a(b-a)(c-b)(d-c)
$$
You can easily find $\det (A^5)$ now and even say that
$$
\det A^n = \det \begin{pmatrix}
a^n&*&*&*\\
0&(b-a)^n&*&*\\
0&0&(c-b)^n&*\\
0&0&0&(d-c)^n
\end{pmatrix}
$$
