quantum mechanics - Proof of de Broglie wavelength for electron

According to de Broglie's wave-particle duality, the relation between electron's wavelength and momentum is $\lambda =h/mv$.

The proof of this is given in my textbook as follows:

1. 
   

   De Broglie first used Einstein's famous equation relating matter and energy,

   $$E=mc^2,$$ where $E=$ energy, $m =$ mass, $c =$ speed of light.
   
   


2. 
   

   Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation,

   $$E=h\nu,$$ where $E =$ energy, $h =$ Plank's constant ($6.62607 \times 10^{-34}\:\mathrm{ J\:s}$), $\nu =$ frequency.
   
   


3. 
   

   Since de Broglie believes particles and wave have the same traits, the two energies would be the same:

   $$mc^2=h\nu.$$
   
   



4. 
   

   Because real particles do not travel at the speed of light, de Broglie substituted $v$, velocity, for $c$, the speed of light:

   $$mv^2=h\nu.$$
   
   

I want a direct proof without substituting $v$ for $c$. Is it possible to prove directly $\lambda=h/mv$ without substituting $v$ for $c$ in the equation $\lambda =h/mc$?