The Legendre's formula gives $\alpha$ in
$$p^\alpha || n!$$
where $p$ is a prime number.
To calculate $\nu_p(n!)$ on paper, one should normally find the quotients $q_i$ in these equations by long division and then sum them up:
$$n=p^i q_i+r_i, \quad 0\le r_i \lt p^i$$
So we will divide $n$ by $p^i$'s multiple times.
The variation I stumbled upon is using the previous quotient and divide by $p$. This method of using recursive divisions is indeed relatively faster as the divisor remains small instead of growing exponentially (as it does in the first method). It is basically saying:
$$Q_0:=n$$ $$Q_i=pQ_{i+1}+R_{i+1}, \quad 0 \le R_i \lt p$$
Manipulating them a bit, we get:
$$n=p^i Q_i + \sum_{j=1}^i p^{j-1} R_j$$
As you see, the equations look similar to those of the first method. But I doubt if this variation will give the correct result because I couldn't prove $\forall i; q_i = Q_i$. This is my approach:
If I prove $0 \le \sum_{j=1}^i p^{j-1} R_j \lt p^i$, then $q_i = Q_i$ and the two methods are equivalent.
$$0\le R_i \lt p$$ $$0\le p^{i-1}R_i \lt p^i$$ $$0\le\sum_{i=1}^k p^{i-1} R_i \lt \sum_{i=1}^k p^i=p\frac{p^k-1}{p-1}$$
I should just check if $p\frac{p^k-1}{p-1} \le p^k$:
$$p(p^k-1) \le p^k (p-1)$$ $$-p \le -p^k$$ $$p \ge p^k$$
Which is wrong since prime numbers are greater than one.
Because of the essence of the proof I used, there might be another approach which works and can prove the two methods are the same.
Is the second method valid?
As Daniel pointed out in comments, your question boils down to this property which I will explain a little more.
If $y=\bigg\lfloor\dfrac{n}{p}\bigg\rfloor$ then $y$ is an integer and $yp \leqslant n \leqslant (y+1)p$
so, since $yp$ is an integer, we have $yp \leqslant \lfloor n \rfloor \leqslant n \leqslant (y+1)p$
and so $\bigg\lfloor\dfrac{\lfloor n \rfloor }{p}\bigg\rfloor = y = \bigg\lfloor\dfrac{n}{p}\bigg\rfloor$.