In the prime integer factorization of $n!$, if $p$ is a prime number and $p$ is a number in this factorization, it's exponent is in the form: $\alpha =\lfloor \frac{n}{p}\rfloor +\lfloor \frac{n}{p^2}\rfloor +\lfloor \frac{n}{p^3}\rfloor +\ldots $.
My thoughts were:
$n!=n(n-1)\ldots 3\cdot 2 $
$n=\sum_{i=1}^\infty p_i ^{\alpha_{n_i}}, (n-1)=\sum_{i=1}^\infty p_i ^{\alpha_{(n-1)_i}}, \ldots \Rightarrow n!=\sum_{i=1}^\infty p_i ^{\alpha_{2_i}+\alpha_{2_i}+\alpha_{3_i}+\ldots+\alpha_{n_i}}$
Therefore I now "only" have to show that $\alpha_j=\lfloor \frac{n}{p_j}\rfloor +\lfloor \frac{n}{p_j^2}\rfloor +\lfloor \frac{n}{p_j^3}\rfloor +\ldots =\alpha_{2_j}+\alpha_{2_j}+\alpha_{3_j}+\ldots+\alpha_{n_j}$
My thoughts on that were to show it by proving that the inequalities in both directions do not apply, and therefore equality must be truth, but I don't really know how to do it.
Hint: $\lfloor n/p \rfloor$ is the number of multiples of $p$ from $1$ to $n$.