Let $X(t)$ be a Poisson process with rate $\lambda$ and $W_k$ be the $k$-th waiting time and $U_k$ be iid uniform random variables on $[0,t]$. Prove: $$E\left[ \sum_{k = 1}^{n}e^{-\beta W_k} | X(t) = n \right] = E\left[ \sum_{k = 1}^{n}e^{-\beta U_k} \right].$$
I can see that $$E\left[ \sum_{k = 1}^{n}e^{-\beta W_k} | X(t) = n \right] = \sum_{w_1, \dots, w_n} \left ( \sum_{k=1}^{n}e^{-\beta w_k} \right )P(W_1 = w_1, \dots, W_n = w_n | X(t) = n)$$ and since $$P(W_1 = w_1, \dots, W_n = w_n | X(t) = n) = n!t^{-n} = P(U_{(1)} = w_1, \dots, U_{(n)} = w_n)$$
we have $$\sum_{w_1, \dots, w_n} \left ( \sum_{k=1}^{n}e^{-\beta w_k} \right )P(W_1 = w_1, \dots, W_n = w_n | X(t) = n) = \sum_{w_1, \dots, w_n} \left ( \sum_{k=1}^{n}e^{-\beta w_k} \right )n!t^{-n} $$ but
$$\sum_{w_1, \dots, w_n} \left ( \sum_{k=1}^{n}e^{-\beta w_k} \right )n!t^{-n} \neq \sum_{w_1, \dots, w_n} \left ( \sum_{k=1}^{n}e^{-\beta w_k} \right )t^{-n} = E\left[ \sum_{k = 1}^{n}e^{-\beta U_k} \right]$$.
Am I miss understanding something here? This seems like a contradiction.