$$\eqalign{ & {a_1} = \sqrt 7 < 7.\,\,{\text{Let }}{a_m} < 7 \cr & {\text{Then }}{a_{m + 1}} = \sqrt {7 + {a_m}} \cr & \Rightarrow \,\,{a^2}_{m + 1} = 7 + {a_m} < 7 + 7 < 14. \cr & \Rightarrow \,\,{a_{m + 1}} < \sqrt {14} < 7; \cr} $$
So by the principle of mathematical induction $${a_n} < 7\,\,\forall \,\,n.$$

Let $$P\left( n \right)$$  be the statement given by $$P\left( n \right):{41^n} - {14^n}$$    is a multiple of 27
For $$n = 1,$$
i.e., $$P\left( 1 \right) = {41^1} - {14^1} = 27 = 1 \times 27,$$
which is a multiple of 27.
$$\therefore P\left( 1 \right)$$  is true.
Let $$P\left( k \right)$$  be true, i.e., $${41^k} - {14^k} = 27\lambda \,\,\,\,\,.....\left( {\text{i}} \right)$$
For $$n = k + 1,$$
$$\eqalign{ & {41^{k + 1}} - {14^{k + 1}} = {41^k}41 - {14^k}14 \cr & = \left( {27\lambda + {{14}^k}} \right)41 - {14^k}14\,\,\,\left[ {{\text{using}}\left( {\text{i}} \right)} \right] \cr & = \left( {27\lambda \times 41} \right) + \left( {{{14}^k} \times 41} \right) - \left( {{{14}^k} \times 14} \right) \cr & = \left( {27\lambda \times 41} \right) + {14^k}\left( {41 - 14} \right) \cr & = \left( {27\lambda \times 41} \right) + \left( {{{14}^k} \times 27} \right) \cr & = 27\left( {41\lambda + {{14}^k}} \right), \cr} $$
which is a multiple of 27.
Therefore, $$P\left( {k + 1} \right)$$   is true when $$P\left( {k} \right)$$  is true.
Hence, from the principle of mathematical induction, the statement is true for all natural numbers $$n.$$