For any random variable $X$ taking values in $\mathbb{N}$, $E[X]= \sum_{n=0}^\infty P(X\gt n)$. In this case, the probability that the sum of the first $n$ numbers is less than $1$ is $1/n!$, the volume of the simplex with vertices at the origin and the standard basis vectors in $n$ dimensions. So, the expected number of draws needed to get a partial sum greater than $1$ is $\sum_{n=0}^\infty 1/n! = e$.

Another approach is to set up a (lag-)differential equation for $f(x)$, the expected number of draws needed for the sum to exceed $x$. For $x \le 0, f(x) = 0$. For $x \gt 0, f(x) = 1+ \int_{x-1}^x f(z) dz$, so $f'(x) = f(x)-f(x-1)$. This has the solution $f(x) = e^x, 0\lt x\le 1$. Another recent question asked about the error from the asymptotic of $f(x) \sim 2x+2/3$.

For any random variable $X$ taking values in $\mathbb{N}$, $E[X]= \sum_{n=0}^\infty P(X\gt n)$. In this case, the probability that the sum of the first $n$ numbers is less than $1$ is $1/n!$, the volume of the simplex with vertices at the origin and the standard basis vectors in $n$ dimensions. So, the expected number of draws needed to get a partial sum greater than $1$ is $\sum_{n=0}^\infty 1/n! = e$.

Another approach is to set up a (lag-)differential equation for $f(x)$, the expected number of draws needed for the sum to exceed $x$. For $x \le 0, f(x) = 0$. For $x \gt 0, f(x) = 1+ \int_{x-1}^x f(z) dz$, so $f'(x) = f(x)-f(x-1)$. This has the solution $f(x) = e^x, 0\lt x\le 1$. Another recent question asked about the error from the asymptotic of $f(x) \sim 2x+2/3$.