Derive the identity elements of lcm and gcd 
Find the identity element of the binary operations $*,*'$ on $\mathbb{N}$ given by $a*b = lcm(a,b)$ and $a*'b = \gcd(a,b)$, where $\mathbb{N}=\{1,2,3...\}$

I know the identity element for lcm is $1$ and there is none for gcd in $\mathbb{N}$. But, how do I prove it mathematically ?
My Attempt
$a*e=a=e*a$
From the fundamental theorem of arithmetic,
$$
a=\prod_ip_i^{a_i}\quad,e=\prod_ip_i^{e_i}\\
lcm(a,b)=\prod_ip_i^{\max(a_i,b_i)}\quad ,\gcd(a,b)=\prod_ip_i^{\min(a_i,b_i)}
$$
For $*:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$
$$
a*e=a\implies lcm(a,e)=\prod_ip_i^{\max(a_i,e_i)}=\prod_ip_i^{a_i}=a\\
\implies \max(a_i,e_i)=a_i\implies e_i\leq a_i \text{ for all }a_i\\
\implies e_i=0\implies \color{red}{e=1}
$$
For $*':\mathbb{N}\times\mathbb{N}\to\mathbb{N}$
$$
a*'e=a\implies \gcd(a,e)=\prod_ip_i^{\min(a_i,e_i)}=\prod_ip_i^{a_i}=a\\
\implies \min(a_i,e_i)=a_i\implies e_i\geq a_i\text{ for all }a_i\\
\implies \text{no }e_i\in\mathbb{N}\text{ satisfies the condition}\implies\color{red}{\text{no identity element for }*'}
$$
Is it the right way to approach the problem ?
 A: In your approach for $\gcd(a,e)=a$, you get that $\min(a_i,e_i)=a_i$ for all $i$. But keep in mind that $e$ will act as an identity for all elements $a$.  
In particular, it should work as identity if we consider $a=e^2$, i.e. we should have $\gcd(e^2,e)=e^2$. But the prime factorization of $e^2$ will of the form $\prod_ip_i^{2e_i}$. So here we have $a_i=2e_i > e_i$. But this violates $\min(2e_i,e_i)=2e_i$. So no such $e$ exists. 
Note: But as suggested by many others, going through prime factorization approach is an overkill for this problem. 
A: $a*b = lcm(a,b)$
Prove that if $lcm(a,e) = a$ for all natural $a$ then $e = 1$.
Pf:  $lcm(a,e) = \frac {ae}{\gcd(a,e)}$.
So if $lcm(a,e) =  \frac {ae}{\gcd(a,e)} =a$ for all natural $a$ then 
$ae = a*\gcd(a,e)$ for all natural numbers.
So $e = \gcd(a,e)$ for all natural numbers.  
So $e$ divides all natural numbers.  
So $e$ divides $1$.  The only natural number that divides $1$ is $1$.
So $e = 1$.
And indeed $a*1 = lcm(a,1) = a$ for all $a$.
$a *' b = \gcd(a,e)$
So prove find an $e$ so that $\gcd(a,e) = a$ for all $a$ or prove one can't exist.
S'pose there were such an $e$ then $\gcd(a,e) =a$ for all possible $a$.
So all $a$ will divide $e$.  So we must find a number that is divisible by any other number.
Well, that would mean $2e|e$ so there is an integer $k$ so that $e = k*(2e)$ so if $e\ne 0$ then $k = \frac 12$ which is not an integer.  So $e = 0$.
And  .... every number does divide $0$ because $a*0 = 0$ for every number.
And $\gcd(a,0) = a$.  Because $a|0$ and $a|a$.  And if $m > a > 0$ then $m\not \mid a$.
But the question now is, was it specified that the identity element must be in $\mathbb N$?
I guess so because otherwize $-1$ would be an identity element for $*$.  SO as $0 \not \in \mathbb N$ I guess we can say there is no natural number that is an identity.
