<s>[INST] From $a_{n+1}=(1+ \frac{1}{n})^n \cdot a_n$ to $a_n$ to be proven by induction 
Find and Prove by induction an explicit formula for $a_n$ if $a_1=1$ and for $n \geq 1$
  $$a_{n+1}=(1+ \frac{1}{n})^n \cdot a_n$$

Checking the pattern:
$$a_1=1$$
$$a_2= 2 \cdot 1$$
$$a_3= (\frac{3}{2})^2 \cdot 2 \cdot 1$$
$$a_4= (\frac{4}{3})^3 \cdot (\frac{3}{2})^2 \cdot 2 \cdot 1$$
$$a_5=  (\frac{5}{4})^4 \cdot   (\frac{4}{3})^3 \cdot (\frac{3}{2})^2 \cdot 2 \cdot 1$$
I can see the general pattern in the index as $(...)^{(n-1)!}$
and I am tempted to state that it is $( \frac{n!}{(n-1)!})^{(n-1)!}
$
Should this be correct, I do not see the factorial manipulation that allow me to write this.
Otherwise, what would the formula be? What would the approach be?
Much appreciated
 [/INST] $a_5=  (\frac{5}{4})^4 \cdot   (\frac{4}{3})^3 \cdot (\frac{3}{2})^2 \cdot 2 \cdot 1
$
$\begin{array}\\
a_n
&=\prod_{k=2}^n (\frac{k}{k-1})^{k-1}\\
&=\dfrac{\prod_{k=2}^n k^{k-1}}{\prod_{k=2}^n (k-1)^{k-1}}\\
&=\dfrac{\prod_{k=2}^n k^{k-1}}{\prod_{k=1}^{n-1} k^k}\\
&=\dfrac{\prod_{k=2}^n k^{k-1}}{\prod_{k=1}^{n-1} kk^{k-1}}\\
&=\dfrac{\prod_{k=2}^n k^{k-1}}{\prod_{k=1}^{n-1} k\prod_{k=1}^{n-1} k^{k-1}}\\
&=\dfrac{n^{n-1}\prod_{k=2}^{n-1} k^{k-1}}{\prod_{k=1}^{n-1} k\prod_{k=2}^{n-1} k^{k-1}}\\
&=\dfrac{n^{n-1}}{(n-1)!}\\
&=\dfrac{n^{n}}{n!}\\
\end{array}
$
</s>