**UPDATE:** I initially gave a full solution to this question, but I just realised that this might be a homework problem.  In case it is a homework problem, here are some general hints on solving it:

You have a sequence of continuous IID random variables which means they are [exchangeable](https://en.wikipedia.org/wiki/Exchangeable_random_variables).  What does this imply about the probability of getting a particular order for the first $n$ values?  Based on this, what is the probability of getting an increasing order for the first $n$ values?  It is possible to figure this out without integrating over the distribution of the underlying random variables.

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Here is the full solution I gave initially (**don't look if you are supposed to figure this out yourself**):

>! Let $U_1, U_2, U_3, \cdots \sim \text{IID U}(0,1)$ be your sequence of independent uniform random variables, and let $N \equiv \max \{ n \in \mathbb{N} | U_1 < U_2 < \cdots < U_n \}$ be the number of increasing elements at the start of the sequence.  Because these are continuous exchangeable random variables, they are almost surely unequal to each other, and any ordering is equally likely, so we have: $$\mathbb{P}(N \geqslant n) = \mathbb{P}(U_1 < U_2 < \cdots < U_n) = \frac{1}{n!}.$$

Next step:

>!(Note that this result holds for any IID sequence of continuous random variables; they don't have to have a uniform distribution.)  So the random variable $N$ has probability mass function $$p_N(n) = \mathbb{P}(N=n) = \frac{1}{n!} - \frac{1}{(n+1)!} = \frac{n}{(n+1)!}.$$

Next step:

>! You will notice that this result accords with the values you have calculated using integration over the underlying values.  The expected value of $N$ is: $$\mathbb{E}(N) = \sum_{n=1}^\infty \mathbb{P}(N \geqslant n) = \sum_{n=1}^\infty \frac{1}{n!} = e - 1 = 1.718282.$$