Let $u_1=\frac1N$, where $N\in\mathbb{Z^+}$, and $u_n=u_{n-1}+(u_{n-1})^2$. Apparently, the number of terms less than $1$, is approximately $N$. Why?

I was playing with recursively defined sequences, and stumbled upon a curious apparent result.

Let $u_1=\frac1N$, where $N\in\mathbb{Z^+}$, and $u_n=u_{n-1}+(u_{n-1})^2$.

Let $f(N)=$ number of terms in the sequence $\{u_n\}$ that are less than $1$. I used Excel to find that:

$f(2)=2$
$f(10)=12$
$f(100)=104$
$f(1000)=1006$
$f(10000)=10008$
$f(100000)=100011$

So it seems that the number of number of terms less than $1$, is approximately $N$. Why?

To put this more precisely,

Prove that $\lim\limits_{N\to\infty}\frac{f(N)}{N}=1$.

My attempt

I found a question called "What's known about recurrences involving $(a_n)^2$?", which led me to the concept of quadratic map, but these did not help me.

I found the sequence A122888 (a large list is here), which gives the coefficients of $x^k$ in the $n$th iteration of $x+x^2$. So for example, terms $a_7$ to $a_{14}$ are $\color{red}{1,3,6,9,10,8,4,1}$. This tells us that, in my original sequence, $u_4=\color{red}{1}(u_1)^1+\color{red}{3}(u_1)^2+\color{red}{6}(u_1)^3+\color{red}{9}(u_1)^4+\color{red}{10}(u_1)^5+\color{red}{8}(u_1)^6+\color{red}{4}(u_1)^7+\color{red}{1}(u_1)^8$. But I do not know what to make of this.