Try for $\lambda=-1$
$$ \det(I-\lambda K) = \left[
\sum_{n=0}^\infty (-\lambda)^n \operatorname{Tr } \Lambda^n(K) \right]=
\exp{ \left( \sum_{n=0}^\infty(-1)^{n+1}\frac{\operatorname{Tr} K^n}{n} (-\lambda)^n \right)}$$
where $\Lambda^n (K)$ is the $n$th exterior power of $K$.
copied from
http://en.wikipedia.org/wiki/Fredholm_determinant
Note that $I-K$ can be put as $D^+ + D^-$, where $D^+$ are strict upper triangular and $D^-$ are strict lower diagonal. Then use the binomial theorem for $(D^+ + D^-)^n$.
Note that $(D^-)^k$ and $(D^+)^k$ are fairly simple to compute;)
We can simply calculate the determinant of an opposite (lower) triangular matrix:
Let $J_n$ be the $n \times n$ matrix with $1$ on the anti-diagonal and $0$ otherwise (i.e. $J_ne_i = e_{n+1-i}$ for all $1 \leq i \leq n$, where $e_1, \dotsc, e_n$ denotes the standard basis). Given any $m \times n$-matrix $A$ the matrix $AJ_n$ originates from $A$ by vertically mirroring its colums from the middle, i.e. swapping the first column with the last, the second with the second last, etc.
If $A$ is an $n \times n$ square matrix then we get from $J_n^2 = I_n$ that
$$
\det(A) = \det(J_n) \det(AJ_n).
$$
In the case of $D_n$ we get that $D_n J_n$ is the $a_n$-scalar multiple a lower triangular matrix with diagonal entries $x-a_1, x-a_2, \dotsc, x-a_{n-1}, 1$, so
$$
\det(D_n)
= \det(J_n) \det(D_n J_n)
= \det(J_n) a_n (x-a_1) \dotsm (x-a_{n-1}).
$$
So the only difference is that we need to know $\det(J_n)$. Because $J_n$ is a permutation matrix, corresponding to $\sigma_n \in S_n$ with $\sigma(i) = n+1-i$, we have $\det(J_n) = \mathrm{sgn}(\sigma_n)$. Notice that
\begin{align*}
\sigma_{2n} &= (1 \;\; 2n) (2 \;\; n-1) \dotsm (n \;\; n+1) \\
\sigma_{2n+1} &= (1 \;\; 2n+1) (2 \;\; n-1) \dotsm (n \;\; n+2).
\end{align*}
So we can just count the number of transpositions used and get that
$$
\mathrm{sgn}(\sigma_n) =
\begin{cases}
\phantom{-}1 & \text{if $n \equiv 0,1 \bmod 4$}, \\
-1 & \text{if $n \equiv 2,3 \bmod 4$},
\end{cases}
= (-1)^{n(n-1)/2}.
$$
So alltogether we have
$$
\det(D_n) = (-1)^{n(n-1)/2} a_n (x-a_1) \dotsm (x-a_{n-1}).
$$
(The nice thing about this is that now that we have calculated $\det(J_n) = (-1)^{n(n-1)/2}$ we can use this to calculate the determinant of opposite triangular and opposite diagonal matrices more ore less in the usual way.)
Best Answer
Hint: the matrix $M = e e^T$ (where $e$ is a column vector consisting of $n$ $1$'s) satisfies $M^2 = n M$, so its eigenvalues are ...