+) With n = 1 then 

\(16^n-15n-1=16-15-1=0⋮225\)

+) With n = 2 then 

\(16^n-15n-1=16^2-15.2-1=225⋮225\)

Suppose \(\left(16^n-15n-1\right)⋮225\)

We are going to prove that \(\left(16^{n+1}-15\left(n+1\right)-1\right)⋮225\) too

We can see \(16^{n+1}-15\left(n+1\right)-1=16.16^n-15n-15-1=\left(16^n-15n-1\right)+15.16^n-15\)

Following Inductive hypothesis , \(\left(16^n-15n-1\right)⋮225\)

And \(15.16^k-15=15\left(16^k-1\right)⋮15.15=225\)

So \(\left(16^{n+1}-15\left(n+1\right)-1\right)⋮225\)

Conclude \(\left(16^n-15n-1\right)⋮225\) \(\forall n\in N\)

Put B = 16ⁿ - 15n - 1 

* With n = 0 <=> B₀ = 16⁰ - 15.0 - 1 = 0 ⋮ 225 

* With n = 1 <=> B₁ = 16¹ - 15.1 - 1 = 0 ⋮ 225

..........................

* With n = k <=> B_k = 16^k - 15k - 1 ⋮ 225

We must prove that with n = k+1 then B ⋮ 225 too

We have: n = k+1

B_k+1 = 16^k+1 - 15.(k+1) - 1 = 16^k.16 - 15k - 15 - 1 = 16^k.16 - 15k - 16 

= 16(16^k - 1) - 15k  

So B_k+1 - 225k = 16(16^k - 15k - 1) = 16.B_k

B_k ⋮ 225, 225 ⋮ 225 so B_k+1 ⋮ 225.

So 16ⁿ - 15n - 1 ⋮ 225