Answer
Let $ P (n)$ be the proposition that:
$$
1 \cdot 1 !+2 \cdot 2 !+\cdots+n \cdot n !=(n+1) !-1
$$
BASIS STEP: P(1) is true, because
by substituting $n=1 $ in the proposition $ P (n)$ , we obtain that $P (1)$ is the statement
$$
1 \cdot 1 ! = (1+1) !-1 =2-1=1
$$
(The left-hand side of this equation is 1 because 1 is the sum of the first positive integer. The right-hand side is found by substituting 1 for $n$ in $(n+1) !-1$)
INDUCTIVE STEP:
For the inductive hypothesis we assume that P(k) holds for an arbitrary
positive integer $k$. That is, we assume that:
$$
1 \cdot 1 !+2 \cdot 2 !+\cdots+k \cdot k !=(k+1) !-1
$$
Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that
$$
1 \cdot 1 !+2 \cdot 2 !+\cdots+(k+1) \cdot (k+1) !=(k+2) !-1
$$
is also true. When we add $(k+1) \cdot (k+1) ! $ to both sides of the equation in $P(k)$, we obtain
$$
\begin{split}
1 \cdot 1 !+2 \cdot 2 !+\cdots+k \cdot k ! + (k+1) \cdot (k+1) ! = \\
\quad[\text { by the inductive hypothesis }] \\
=\left[(k+1) !-1\right]+(k+1) \cdot (k+1) ! \\
=(1+k+1) \cdot (k+1) ! -1\\
=(k+2) \cdot (k+1) ! -1\\
=(k+2) ! -1\\
\end{split}
$$
We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for all positive integers $n$.
That is, we have proven that
$$
1 \cdot 1 !+2 \cdot 2 !+\cdots+n \cdot n !=(n+1) !-1
$$
for all positive integers $n$.
Work Step by Step
Let $ P (n)$ be the proposition that:
$$
1 \cdot 1 !+2 \cdot 2 !+\cdots+n \cdot n !=(n+1) !-1
$$
BASIS STEP: P(1) is true, because
by substituting $n=1 $ in the proposition $ P (n)$ , we obtain that $P (1)$ is the statement
$$
1 \cdot 1 ! = (1+1) !-1 =2-1=1
$$
(The left-hand side of this equation is 1 because 1 is the sum of the first positive integer. The right-hand side is found by substituting 1 for $n$ in $(n+1) !-1$)
INDUCTIVE STEP:
For the inductive hypothesis we assume that P(k) holds for an arbitrary
positive integer $k$. That is, we assume that:
$$
1 \cdot 1 !+2 \cdot 2 !+\cdots+k \cdot k !=(k+1) !-1
$$
Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that
$$
1 \cdot 1 !+2 \cdot 2 !+\cdots+(k+1) \cdot (k+1) !=(k+2) !-1
$$
is also true. When we add $(k+1) \cdot (k+1) ! $ to both sides of the equation in $P(k)$, we obtain
$$
\begin{split}
1 \cdot 1 !+2 \cdot 2 !+\cdots+k \cdot k ! + (k+1) \cdot (k+1) ! = \\
\quad[\text { by the inductive hypothesis }] \\
=\left[(k+1) !-1\right]+(k+1) \cdot (k+1) ! \\
=(1+k+1) \cdot (k+1) ! -1\\
=(k+2) \cdot (k+1) ! -1\\
=(k+2) ! -1\\
\end{split}
$$
We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for all positive integers $n$.
That is, we have proven that
$$
1 \cdot 1 !+2 \cdot 2 !+\cdots+n \cdot n !=(n+1) !-1
$$
for all positive integers $n$.
