This is an interview question for a quantitative analyst position, reported [here](https://www.glassdoor.com/Interview/Assume-X-i-i-i-d-Unif-0-1-What-is-the-expected-length-of-a-sequence-that-is-monotonically-increasing-when-drawn-from-t-QTN_2162387.htm).  Suppose we are drawing from a uniform $[0,1]$ distribution and the draws are iid, what is the expected length of a monotonically increasing distribution? I.e., we stop drawing if the current draw is smaller than or equal to the previous draw.

I've gotten the first few:
$$
        \Pr(\text{length} = 1) = \int_0^1 \int_0^{x_1} \mathrm{d}x_2\,  \mathrm{d}x_1 = 1/2 
$$
$$
        \Pr(\text{length} = 2) = \int_0^1 \int_{x_1}^1 \int_0^{x_2}  \mathrm{d}x_3 \, \mathrm{d}x_2 \, \mathrm{d}x_1 = 1/3 
$$
$$
         \Pr(\text{length} = 3) = \int_0^1 \int_{x_1}^1 \int_{x_2}^1 \int_0^{x_3}  \mathrm{d}x_4\,  \mathrm{d}x_3\,  \mathrm{d}x_2\,  \mathrm{d}x_1 = 1/8 
$$

but I find calculating these nested integrals increasingly difficult and I'm not getting the "trick" to generalize to 
$$
         \Pr(\text{length} = n)
$$

I know the final answer is structured
$$
        \mathbb E(\text{length of a monotonically increasing sequence}) = \sum_{n=1}^{\infty}n\Pr(\text{length} = n) 
$$

Any ideas on how to answer this question?