a.
[tex]\begin{cases}c_0=5\\c_n={c_{n-1}}^2&\text{for }n\ge1\end{cases}[/tex]
For [tex]n=1[/tex], we have by this definition [tex]c_1={c_0}^2=5^2[/tex] which is equal to [tex]5^{2^n}[/tex] for [tex]n=1[/tex].
Assume [tex]c_n=5^{2^n}[/tex] for some [tex]n=k[/tex]. We want to use this to show that this implies [tex]c_{k+1}=5^{2^{k+1}}[/tex]. By the recursive definition,
[tex]c_{k+1}={c_k}^2=(5^{2^k})^2=5^{2^{k+1}}[/tex]
and we're done.
b.
[tex]\begin{cases}b_0=1\\b_n=2b_{n-1}+1&\text{for }n\ge1\end{cases}[/tex]
For [tex]n=0[/tex], we have [tex]b_0=1[/tex], and [tex]2^{n+1}-1=1[/tex].
Assume [tex]b_n=2^{n+1}-1[/tex] for some [tex]n=k[/tex]. Then for [tex]n=k+1[/tex], we have
[tex]b_{k+1}=2b_k+1=2\left(2^{k+1}-1\right)+1=2^{k+2}-1[/tex]
and we're done.